8b. Little Billie weighs 360 N on Earth. What is Little Billie's mass on the moon where the force of gravity is approximately 1/6-th that of Earth's? ________ Explain or show your work.
Answer: ~36 kg
The mass of an object is related to weight by the equation W = m•g where g = ~10 m/s/s on Earth and one-sixth this value (~1.67) on the moon. So if Billy weighs 360 N on Earth, then his mass is approximately ~36 kg. His mass on the moon will be the same as his mass on Earth. Only his weight changes when on the moon; rather than being 360 N, it is 60 N. His weight on the moon could be found by multiplying his mass by the value of g on the moon: (36 kg) • (9.8/6 m/s/s) = ~60 N
9. TRUE or FALSE:
An object which is moving rightward has a rightward force acting upon it.
Answer: False
An object which is accelerating rightward must have a rightward force and a rightward net force acting upon it. But an object which is merely moving rightward does not necessarily have a rightward force upon it. A car that is moving rightward and skidding to a stop would not have a rightward force acting upon it.
10. The amount of net force required to keep a 5-kg object moving rightward with a constant velocity of 2 m/s is ____.
a. 0 N | b. 0.4 N | c. 2 N | d. 2.5 N | e. 5 N |
Net force is always m•a. In this case, the velocity is constant so the acceleration is zero and the net force is zero. Constant velocity motion can always be associated with a zero net force.
11. TRUE or FALSE:
For an object resting upon a non-accelerating surface, the normal force is equal to the weight of the object.
Quite surprisingly to many, the normal force is not necessarily always equal to the weight of an object. Suppose that a person weighs 800 N and sits at rest upon a table. Then suppose another person comes along and pushes downwards upon the persons shoulders, applying a downward force of 200 N. With the additional downward force of 200 N acting upon the person, the total upward force must be 1000 N. The normal force supplies the upward force to support both the force of gravity and the applied force acting upon the person. Its value is equal to 1000 N which is not the same as the force of gravity of the person.
12. Which one(s) of the following force diagrams depict an object moving to the right with a constant speed? List all that apply.
If an object is moving at a constant speed in a constant rightward direction, then the acceleration is zero and the net force must be zero. Choice B and D show a rightward net force and therefore a rightward acceleration, inconsistent with the described motion.
13. According to Newton's third law, every force is accompanied by an equal and opposite reaction force. The reason that these forces do not cancel each other is ____.
a. the action force acts for a longer time period b. the two forces are not always in the same direction c. one of the two forces is greater than the other d. the two forces act upon different objects; only forces on the same object can balance each other. e. ... nonsense! They do cancel each other. Objects accelerate because of the presence of a third force.
Action and reaction forces always act upon the interacting objects for the same amount of time with the same magnitude. So if object A pushes on object B, then object B simultaneously pushes on object A with the same amount of force. The force on object B will be one of perhaps many forces which will govern its motion. But the reaction force is on object A and cannot contribute to object B's motion since it is not acting upon object B. Action-reaction forces can NEVER cancel each other.
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14. As you sit in your chair and study your physics (presuming that you do), the force of gravity acts downward upon your body. The reaction force to the force of the Earth pulling you downward is ___.
a. the force of the chair pushing you upward
b. the force of the floor pushing your chair upward
c. the force of the Earth pushing you upward
d. the force of air molecules pushing you upwards
e. the force of your body pulling the Earth upwards
f. ... nonsense! Gravity is a field force and there is no such reaction force.
The most common wrong answer is a - the force of the chair pushing you upward. As you sit in your chair, the chair is indeed pushing you upward but this is not the reaction force to the force of the Earth pulling you downward. The chair pushing you upward is the reaction force to you sitting on it and pushing the chair downward. To determine the action-reaction force pairs if given a statement of the form object A pulls X-ward on object B , simply take the subject and the object in the sentence and switch their places and then change the direction to the opposite direction (so the reaction force is object B pulls object A in the opposite direction of X). So if the Earth pulls you down ward, then the reaction force is you pull the Earth up ward.
15. A golf pro places a ball at rest on the tee, lines up his shot, draws back his club, and lets one rip. During the contact of the golf club with the golf ball, the force of the club on the ball is ____ the force of the ball on the club and the acceleration of the club is ____ than the acceleration of the ball.
a. greater than, greater than | b. greater than, equal to | c. greater than, less than |
d. less than, less than | e. less than, equal to | f. less than, greater than |
g. equal to, equal to | h. equal to, greater than | i. equal to, less than |
For every action, there is an equal and opposite reaction force. In this case, the force on the club is equal to the force on the ball. The subsequent accelerations of the interacting objects will be inversely dependent upon mass. The more massive club will have less acceleration than the less massive ball.
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Each one of Newton's Laws can play a role in any one particular situation. However, one of the laws is often most obviously dominant in governing the motion of a situation. Pick which of Newton's most governs the situations described below.
a. First Law (inertia) | b. Second Law (F = m•a) | c. Third Law (action-reaction) |
16. A helicopter must have two sets of blades in order to fly with stability.
17. If you were in an elevator and the cable broke, jumping up just before the elevator hit the ground would not save you. Sorry.
18. You usually jerk a paper towel from a roll in order to tear it instead of pulling it smoothly.
19. A student desk changes the amount of force it puts on other objects throughout a school day.
20. Heavy objects are not easier to move around in a horizontal fashion on the Moon than on the Earth.
21. The stronger, heavier team in a tug-of-war does not create a larger tension in the rope than the weaker, lighter team.
Answers: See answers and explanations below.
16. C - As the helicopter blades spin and push air in one direction, the air pushes the blades in the opposite direction; the result is that the helicopter can begin to rotate about the axis of the blade. To counteract this rotation, a second set of blades is required.
17. A - An object moving downwards will continue to move downwards unless acted upon by an unbalanced force. If you make an effort to supply such a force in an attempt to suddenly alter the direction of your motion, then you are creating a greater velocity change than if you merely hit the ground and stopped. If this greater velocity change occurred suddenly (in the same amount of time as the stopping of you and the elevator), then you would experience a greater acceleration, a greater net force, and a greater ouch mark than if you had merely hit the ground and stopped.
18. A - The paper towel is at rest and resists changes in its at rest state. So if you apply a sudden force to one of the paper towel sheets, the great mass of the remainder of the roll will resist a change in its at rest state and the roll will easily break at the perforation.
19. C - As a student sits in the seat, they are applying a downward force upon the seat. The reaction force is that the seat applies an upward force upon the person. A weightier person will apply more downward force than a lighter person. Thus, the seat will constantly be changing the amount of reaction force throughout the day as students of different weight sit in it.
20. A - All objects have inertia or a tendency to resist changes in their state of motion. This inertia is dependent solely upon mass and is subsequently not altered by changes in the gravitational environment. To move an object horizontally, one must apply a force; this force will be resisted by the mass or inertia of the object. On the moon, the object offers the same amount of inertia as on Earth; it is just as difficult (or easy) to move around.
21. C - A rope encounters tension when pulled on at both ends. The tension in the rope is everywhere the same. If team A were to pull at the left end, then the left end would pull back with the same amount of force upon team A. This force is the same everywhere in the rope, including at the end where team B is pulling. Thus team B is pulling back on the rope with the same force as team A. So if the forces are the same at each end, then how can a team ever win a tug-of-war. The way a stronger team wins a tug-of-war is with their legs. They push upon the ground with a greater force than the other team. This force upon the ground results in a force back upon the team in order for them to pull the rope and the other team backwards across the line.
For the next several questions, consider the velocity-time plot below for the motion of an object along a horizontal surface. The motion is divided into several time intervals, each labeled with a letter.
22. During which time interval(s), if any, are there no forces acting upon the object? List all that apply.
23. During which time interval(s), if any, are the forces acting upon the object balanced? List all that apply.
24. During which time interval(s), if any, is there a net force acting upon the object? List all that apply.
25. During which time interval(s), if any, is the net force acting upon the object directed toward the right? List all that apply.
26. During which time interval(s), if any, is the net force acting upon the object directed toward the left? List all that apply.
22. None - If an object is on a surface, one can be guaranteed of at least two forces - gravity and normal force.
23. BDFH - If the forces are balanced, then an object is moving with a constant velocity. This is represented by a horizontal line on a velocity-time plot.
24. ACEG - If an object has a net force upon it, then it is accelerating. Acceleration is represented by a sloped line on a velocity-time plot.
25. AE - If the net force is directed to the right, then the acceleration is to the right (in the + direction). This is represented by a line with a + slope (i.e., upward slope).
26. CG - If the net force is directed to the left, then the acceleration is to the left (in the - direction). This is represented by a line with a - slope (i.e., downward slope).
For the next several questions, consider the dot diagram below for the motion of an object along a horizontal surface. The motion is divided into several time intervals, each labeled with a letter.
27. During which time interval(s), if any, are there no forces acting upon the object? List all that apply.
28. During which time interval(s), if any, are the forces acting upon the object balanced? List all that apply.
29. During which time interval(s), if any, is there a net force acting upon the object? List all that apply.
30. During which time interval(s), if any, is the net force acting upon the object directed toward the right? List all that apply.
31. During which time interval(s), if any, is the net force acting upon the object directed toward the left? List all that apply.
27. None - If an object is on a surface, one can be guaranteed of at least two forces - gravity and normal force.
28. ACEGI - If the forces are balanced, then an object is moving with a constant velocity or at rest. This is represented by a section of a dot diagram where the dots are equally spaced apart (moving with a constant velocity) or not even spaced apart at all (at rest).
29. BDFH - If an object has a net force upon it, then it is accelerating. Acceleration is represented by a section of a dot diagram in which the spacing between consecutive dots is either increasing or decreasing.
30. BF - If the net force is directed to the right, then the acceleration is to the right (in the + direction). This is represented by a dot diagram in which the dots are increasing their separation distance as the object moves from left to right.
31. DH - If the net force is directed to the left, then the acceleration is to the left (in the - direction). This is represented by a dot diagram in which the dots are decreasing their separation distance as the object moves from left to right.
For the next several questions, consider the trajectory diagram shown below for a projectile thrown at an angle to the horizontal. The vector arrows represent the horizontal and vertical components of the projectile's velocity . Several points in the trajectory are labeled with a letter. Use the trajectory diagram to answer the next several questions. (Consider air resistance to be negligible.)
32. At which point(s), if any, are there no forces acting upon the object? List all that apply.
33. At which point(s), if any, are the forces acting upon the object balanced? List all that apply.
34. At which point(s), if any, is there a net force acting upon the object? List all that apply.
35. At which point(s), if any, is the net force acting upon the object directed toward the right? List all that apply.
36. At which point(s), if any, is the net force acting upon the object directed upward? List all that apply.
Answers: See answers and explanations below
34. ABCDEFG
This object is a projectile as can be seen by its constant horizontal velocity and a changing vertical velocity (besides that, the problem states that this is a projectile). A projectile is an object upon which the only force is gravity. Gravity acts downward to accelerate an object downward. This force is an unbalanced force or net force. It causes a vertical rising object to slow down and a falling object to speed up.
The presence of a horizontal velocity does not demand a horizontal force, only a balance of horizontal forces. Having no forces horizontally would cause the projectile to move at a constant horizontal speed once it is launched. Similarly an upward force is not needed on this projectile. When launched, an upward velocity is imparted to it; this velocity is steadily decreased as the object is acted upon by the downward force of gravity. An upward force would only be required for an object which is speeding up as it rises upward.
Solved problems in Newton’s laws of motion – Newton’s second law of motion
Wanted : net force (∑F)
2. Mass of an object = 1 kg, net force ∑F = 2 Newton. Determine the magnitude and direction of the object’s acceleration….
3. Object’s mass = 2 kg, F 1 = 5 Newton, F 2 = 3 Newton. The magnitude and direction of the acceleration is…
Mass (m) = 2 kg
The magnitude of the acceleration :
Wanted : The magnitude and direction of the acceleration (a)
∑ F = F 1x – F 2 = 5 – 1 = 4 Newton
Force (F) = 200 N
∑ F = net force, m = mass, a = acceleration
200 – F g = 120
Mass of block B (m B ) = 300 gram = 0.3 kg
Apply Newton’s second law of motion on both blocks :
5 – 4 = (0.4) a
N A – w A = m A a
8. An object with weight of 4 N supported by a cord and pulley. A force of 2 N acts on the block and one end of the cord pulled by a force of 9 N. Determine the net force acts on object X.
Weight of X (w X ) = 4 Newton
The tension force has the same magnitude in all part of the cord. So the tension force is 9 N.
The correct answer is A.
B. 1.70 N and its direction same as force acted by Andrew
F 1 + F 2 = m a
Determine the object’s acceleration.
13. Which statements below describes Newton’s third law?
Leave a comment cancel reply.
Learning objectives.
By the end of this section, you will be able to:
Success in problem solving is necessary to understand and apply physical principles. We developed a pattern of analyzing and setting up the solutions to problems involving Newton’s laws in Newton’s Laws of Motion ; in this chapter, we continue to discuss these strategies and apply a step-by-step process.
We follow here the basics of problem solving presented earlier in this text, but we emphasize specific strategies that are useful in applying Newton’s laws of motion . Once you identify the physical principles involved in the problem and determine that they include Newton’s laws of motion, you can apply these steps to find a solution. These techniques also reinforce concepts that are useful in many other areas of physics. Many problem-solving strategies are stated outright in the worked examples, so the following techniques should reinforce skills you have already begun to develop.
Applying newton’s laws of motion.
Let’s apply this problem-solving strategy to the challenge of lifting a grand piano into a second-story apartment. Once we have determined that Newton’s laws of motion are involved (if the problem involves forces), it is particularly important to draw a careful sketch of the situation. Such a sketch is shown in Figure 6.2 (a). Then, as in Figure 6.2 (b), we can represent all forces with arrows. Whenever sufficient information exists, it is best to label these arrows carefully and make the length and direction of each correspond to the represented force.
As with most problems, we next need to identify what needs to be determined and what is known or can be inferred from the problem as stated, that is, make a list of knowns and unknowns. It is particularly crucial to identify the system of interest, since Newton’s second law involves only external forces. We can then determine which forces are external and which are internal, a necessary step to employ Newton’s second law. (See Figure 6.2 (c).) Newton’s third law may be used to identify whether forces are exerted between components of a system (internal) or between the system and something outside (external). As illustrated in Newton’s Laws of Motion , the system of interest depends on the question we need to answer. Only forces are shown in free-body diagrams, not acceleration or velocity. We have drawn several free-body diagrams in previous worked examples. Figure 6.2 (c) shows a free-body diagram for the system of interest. Note that no internal forces are shown in a free-body diagram.
Once a free-body diagram is drawn, we apply Newton’s second law. This is done in Figure 6.2 (d) for a particular situation. In general, once external forces are clearly identified in free-body diagrams, it should be a straightforward task to put them into equation form and solve for the unknown, as done in all previous examples. If the problem is one-dimensional—that is, if all forces are parallel—then the forces can be handled algebraically. If the problem is two-dimensional, then it must be broken down into a pair of one-dimensional problems. We do this by projecting the force vectors onto a set of axes chosen for convenience. As seen in previous examples, the choice of axes can simplify the problem. For example, when an incline is involved, a set of axes with one axis parallel to the incline and one perpendicular to it is most convenient. It is almost always convenient to make one axis parallel to the direction of motion, if this is known. Generally, just write Newton’s second law in components along the different directions. Then, you have the following equations:
(If, for example, the system is accelerating horizontally, then you can then set a y = 0 . a y = 0 . ) We need this information to determine unknown forces acting on a system.
As always, we must check the solution. In some cases, it is easy to tell whether the solution is reasonable. For example, it is reasonable to find that friction causes an object to slide down an incline more slowly than when no friction exists. In practice, intuition develops gradually through problem solving; with experience, it becomes progressively easier to judge whether an answer is reasonable. Another way to check a solution is to check the units. If we are solving for force and end up with units of millimeters per second, then we have made a mistake.
There are many interesting applications of Newton’s laws of motion, a few more of which are presented in this section. These serve also to illustrate some further subtleties of physics and to help build problem-solving skills. We look first at problems involving particle equilibrium, which make use of Newton’s first law, and then consider particle acceleration, which involves Newton’s second law.
Recall that a particle in equilibrium is one for which the external forces are balanced. Static equilibrium involves objects at rest, and dynamic equilibrium involves objects in motion without acceleration, but it is important to remember that these conditions are relative. For example, an object may be at rest when viewed from our frame of reference, but the same object would appear to be in motion when viewed by someone moving at a constant velocity. We now make use of the knowledge attained in Newton’s Laws of Motion , regarding the different types of forces and the use of free-body diagrams, to solve additional problems in particle equilibrium .
Different tensions at different angles.
Thus, as you might expect,
This gives us the following relationship:
Note that T 1 T 1 and T 2 T 2 are not equal in this case because the angles on either side are not equal. It is reasonable that T 2 T 2 ends up being greater than T 1 T 1 because it is exerted more vertically than T 1 . T 1 .
Now consider the force components along the vertical or y -axis:
This implies
Substituting the expressions for the vertical components gives
There are two unknowns in this equation, but substituting the expression for T 2 T 2 in terms of T 1 T 1 reduces this to one equation with one unknown:
which yields
Solving this last equation gives the magnitude of T 1 T 1 to be
Finally, we find the magnitude of T 2 T 2 by using the relationship between them, T 2 = 1.225 T 1 T 2 = 1.225 T 1 , found above. Thus we obtain
Particle acceleration.
We have given a variety of examples of particles in equilibrium. We now turn our attention to particle acceleration problems, which are the result of a nonzero net force. Refer again to the steps given at the beginning of this section, and notice how they are applied to the following examples.
Drag force on a barge.
The drag of the water F → D F → D is in the direction opposite to the direction of motion of the boat; this force thus works against F → app , F → app , as shown in the free-body diagram in Figure 6.4 (b). The system of interest here is the barge, since the forces on it are given as well as its acceleration. Because the applied forces are perpendicular, the x - and y -axes are in the same direction as F → 1 F → 1 and F → 2 . F → 2 . The problem quickly becomes a one-dimensional problem along the direction of F → app F → app , since friction is in the direction opposite to F → app . F → app . Our strategy is to find the magnitude and direction of the net applied force F → app F → app and then apply Newton’s second law to solve for the drag force F → D . F → D .
The angle is given by
From Newton’s first law, we know this is the same direction as the acceleration. We also know that F → D F → D is in the opposite direction of F → app , F → app , since it acts to slow down the acceleration. Therefore, the net external force is in the same direction as F → app , F → app , but its magnitude is slightly less than F → app . F → app . The problem is now one-dimensional. From the free-body diagram, we can see that
However, Newton’s second law states that
This can be solved for the magnitude of the drag force of the water F D F D in terms of known quantities:
Substituting known values gives
The direction of F → D F → D has already been determined to be in the direction opposite to F → app , F → app , or at an angle of 53 ° 53 ° south of west.
In Newton’s Laws of Motion , we discussed the normal force , which is a contact force that acts normal to the surface so that an object does not have an acceleration perpendicular to the surface. The bathroom scale is an excellent example of a normal force acting on a body. It provides a quantitative reading of how much it must push upward to support the weight of an object. But can you predict what you would see on the dial of a bathroom scale if you stood on it during an elevator ride? Will you see a value greater than your weight when the elevator starts up? What about when the elevator moves upward at a constant speed? Take a guess before reading the next example.
What does the bathroom scale read in an elevator.
From the free-body diagram, we see that F → net = F → s − w → , F → net = F → s − w → , so we have
Solving for F s F s gives us an equation with only one unknown:
or, because w = m g , w = m g , simply
No assumptions were made about the acceleration, so this solution should be valid for a variety of accelerations in addition to those in this situation. ( Note: We are considering the case when the elevator is accelerating upward. If the elevator is accelerating downward, Newton’s second law becomes F s − w = − m a . F s − w = − m a . )
Thus, the scale reading in the elevator is greater than his 735-N (165-lb.) weight. This means that the scale is pushing up on the person with a force greater than his weight, as it must in order to accelerate him upward. Clearly, the greater the acceleration of the elevator, the greater the scale reading, consistent with what you feel in rapidly accelerating versus slowly accelerating elevators. In Figure 6.5 (b), the scale reading is 735 N, which equals the person’s weight. This is the case whenever the elevator has a constant velocity—moving up, moving down, or stationary.
Now calculate the scale reading when the elevator accelerates downward at a rate of 1.20 m/s 2 . 1.20 m/s 2 .
The solution to the previous example also applies to an elevator accelerating downward, as mentioned. When an elevator accelerates downward, a is negative, and the scale reading is less than the weight of the person. If a constant downward velocity is reached, the scale reading again becomes equal to the person’s weight. If the elevator is in free fall and accelerating downward at g , then the scale reading is zero and the person appears to be weightless.
Two attached blocks.
For block 1: T → + w → 1 + N → = m 1 a → 1 T → + w → 1 + N → = m 1 a → 1
For block 2: T → + w → 2 = m 2 a → 2 . T → + w → 2 = m 2 a → 2 .
Notice that T → T → is the same for both blocks. Since the string and the pulley have negligible mass, and since there is no friction in the pulley, the tension is the same throughout the string. We can now write component equations for each block. All forces are either horizontal or vertical, so we can use the same horizontal/vertical coordinate system for both objects
When block 1 moves to the right, block 2 travels an equal distance downward; thus, a 1 x = − a 2 y . a 1 x = − a 2 y . Writing the common acceleration of the blocks as a = a 1 x = − a 2 y , a = a 1 x = − a 2 y , we now have
From these two equations, we can express a and T in terms of the masses m 1 and m 2 , and g : m 1 and m 2 , and g :
Calculate the acceleration of the system, and the tension in the string, when the masses are m 1 = 5.00 kg m 1 = 5.00 kg and m 2 = 3.00 kg . m 2 = 3.00 kg .
Atwood machine.
Determine a general formula in terms of m 1 , m 2 m 1 , m 2 and g for calculating the tension in the string for the Atwood machine shown above.
Physics is most interesting and most powerful when applied to general situations that involve more than a narrow set of physical principles. Newton’s laws of motion can also be integrated with other concepts that have been discussed previously in this text to solve problems of motion. For example, forces produce accelerations, a topic of kinematics , and hence the relevance of earlier chapters.
When approaching problems that involve various types of forces, acceleration, velocity, and/or position, listing the givens and the quantities to be calculated will allow you to identify the principles involved. Then, you can refer to the chapters that deal with a particular topic and solve the problem using strategies outlined in the text. The following worked example illustrates how the problem-solving strategy given earlier in this chapter, as well as strategies presented in other chapters, is applied to an integrated concept problem.
What force must a soccer player exert to reach top speed.
This is a reasonable result: The acceleration is attainable for an athlete in good condition. The force is about 50 pounds, a reasonable average force.
The soccer player stops after completing the play described above, but now notices that the ball is in position to be stolen. If she now experiences a force of 126 N to attempt to steal the ball, which is 2.00 m away from her, how long will it take her to get to the ball?
What force acts on a model helicopter.
The magnitude of the force is now easily found:
Find the direction of the resultant for the 1.50-kg model helicopter.
Baggage tractor.
Recall that v = d s d t v = d s d t and a = d v d t a = d v d t . If acceleration is a function of time, we can use the calculus forms developed in Motion Along a Straight Line , as shown in this example. However, sometimes acceleration is a function of displacement. In this case, we can derive an important result from these calculus relations. Solving for dt in each, we have d t = d s v d t = d s v and d t = d v a . d t = d v a . Now, equating these expressions, we have d s v = d v a . d s v = d v a . We can rearrange this to obtain a d s = v d v . a d s = v d v .
Motion of a projectile fired vertically.
The acceleration depends on v and is therefore variable. Since a = f ( v ) , a = f ( v ) , we can relate a to v using the rearrangement described above,
We replace ds with dy because we are dealing with the vertical direction,
We now separate the variables ( v ’s and dv ’s on one side; dy on the other):
Thus, h = 114 m . h = 114 m .
If atmospheric resistance is neglected, find the maximum height for the mortar shell. Is calculus required for this solution?
Explore the forces at work in this simulation when you try to push a filing cabinet. Create an applied force and see the resulting frictional force and total force acting on the cabinet. Charts show the forces, position, velocity, and acceleration vs. time. View a free-body diagram of all the forces (including gravitational and normal forces).
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Study concepts, example questions & explanations for ap physics 1, all ap physics 1 resources, example questions, example question #1 : newton's second law.
Use Newton's second law to solve this problem.
When the elevator is not moving, we get
However, when the elevator is accelerating downward, the man appears to be lighter since the elevator is negating some of the force from gravity. Written as an equation, we have:
Putting in our values, we get:
A skydiver of mass 70kg has jumped out of a plane two miles above the surface of the earth. After 20 seconds, he has reached terminal velocity, meaning he is no longer accelerating. What is the force of the air on the skydiver's body?
This question is testing your understanding of terminal velocity and Newton's second law. Since the skydiver is at terminal velocity, the force of the air is equal to the force of gravity, resulting in zero net force and thus no acceleration. We just need to calculate the force of gravity on the skydiver to find the force of the air:
There are two forces in play in this scenario. The first is gravity, and the second is air resistance. Since they are opposing each other, we can write:
Substituting in Newton's second law, we get:
Rearranging for the force of air resistance, we get:
Plugging in our values from the problem statement:
A diver of 50kg jumps from a platform 20m high into a pool. If the diver decelerates at a constant rate to zero velocity in 0.8 seconds after hitting the water, what is the force that the water exerts on the diver?
We can use the equation for conservation of energy to calculate the velocity of the diver as he hits the water:
Cancel out initial kinetic and final potential energies, and plug in our expressions:
Cancel out mass and rearranging for final velocity:
Plug in our values:
We know that the diver then decelerates from this velocity to zero in 0.8 seconds, so we can calculate the acceleration:
Then use Newton's second law to calculate the force on the diver:
The truck experiences the greater force and the greater acceleration
The truck experiences the greater force and the car experiences the greater acceleration
The car experiences the greater force and the greater acceleration
The car and the truck experience equal force and the car experiences greater acceleration
Both the car and the truck experience equal force and acceleration
The car and the truck experience equal and opposite forces, but since the car has a smaller mass it will experience greater acceleration than the truck according to the equation F = ma.
A greater mass will decrease the acceleration.
Since we are neglecting air resistance, there are two forces in play: gravity and friction. Therefore, we can use Newton's second law to write the following:
Substituting in an expression for the force of gravity and rearranging for the force of friction, we get:
Use Newton's second law.
A constant force of 30N acts on a a 10kg box as shown in the diagram. If the box is originally at rest, what will be its velocity after 5s?
The box has a constant force acting on it pulling it towards the left. Therefore we can write this as:
Since the force is constant, this means that is is causing the box to move with a constant acceleration that we can calculate using Newton's second law of motion:
Now that we know the acceleration, we can calculate the final velocity after 5 seconds:
So we have that
Note that the negative sign indicates that the box is moving to the left.
The force applied by the weaker person can be calculated using Newton's second law, which states:
The net force is equal to the product of the mass of the object and the acceleration of the object. We were given the mass and acceleration of the object, but only the ratio of the applied forces:
Is it possible to have a non-zero number of forces acting on an object (of non-zero mass), yet the object doesn't acclerate?
Newton's second law states that the net force, or the vector sum of all the forces acting on an object, equals the mass times the acceleration. So, it is possible to have forces act on an object without acceleration if the forces are oriented such that they vector sum to zero. An example would be a person sitting in a chair. Gravity and the normal force both act on the person. However, these forces are equal in magnitude and opposite in direction. So the person doesn't accelerate.
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Written by Abigail Hadden
Modified & Updated: 27 Jul 2024
Reviewed by Sherman Smith
Khabarovsk is a fascinating city filled with rich history, vibrant culture, and breathtaking natural landscapes. Located in the Far East region of Russia, Khabarovsk is the administrative center of Khabarovsk Krai. With its intriguing past and flourishing present, it is a city that captures the imagination of both locals and tourists alike.
In this article, we will dive into 36 fascinating facts about Khabarovsk that will give you a deeper understanding of this captivating city. From its important role in the Russian Far East to its stunning architectural gems and delicious local cuisine, there is no shortage of interesting tidbits to uncover.
So, buckle up and get ready to explore the hidden gems and lesser-known details about Khabarovsk that will leave you amazed and eager to pack your bags for an adventure in this remarkable city.
Khabarovsk is located on the banks of the Amur River and serves as an important economic and cultural center in the region.
Yerofey Khabarov was a Russian explorer and adventurer who played a key role in the Russian exploration and colonization of the Far East.
The city was named to honor the pioneering spirit and contributions of Yerofey Khabarov.
With a diverse population, the city is home to people of various ethnicities, including Russians, Ukrainians, Koreans , and Chinese.
The city has hot and humid summers, with temperatures reaching as high as 35 degrees Celsius, and cold winters, with temperatures dropping below freezing.
Khabarovsk is the administrative center of the Khabarovsk Krai region and is responsible for governing the surrounding areas.
Khabarovsk is a major stop along the famous Trans-Siberian Railway, connecting Moscow with the Russian Far East, China, and beyond.
The city is surrounded by picturesque mountains, forests, and rivers, offering residents and visitors alike a variety of outdoor activities and breathtaking views.
Spanning the Amur River , the Khabarovsk Bridge is a symbol of the city’s connectivity and serves as a vital transportation link.
The city is known for its manufacturing industries, including machinery, food processing, and timber, as well as its busy river port and international airport.
The city’s multicultural population has contributed to a rich tapestry of traditions, languages, and cuisines from different ethnic groups.
Showcasing the history and culture of the region, the museum houses a vast collection of artifacts and exhibits.
The local ice hockey team , known as the Amur Tigers, competes in the Kontinental Hockey League and has a passionate fan base.
Visitors can enjoy acrobatics, animal shows, and other dazzling acts at the city’s beloved circus.
Named after the Russian revolutionary leader, the stadium hosts various sporting events, including football matches and athletics competitions.
Students come from all over the country to study in Khabarovsk, contributing to the city’s vibrant academic community.
Known for its engineering and technical programs, the university plays a key role in shaping the region’s workforce.
During the winter months, artists create intricate ice sculptures that attract locals and tourists alike.
Music lovers can attend classical concerts and symphony performances at the city’s prestigious philharmonic hall.
The park offers beautiful gardens, walking paths, and amusement rides, providing a peaceful retreat for residents.
From traditional Russian cuisine to international flavors, the city boasts a wide range of restaurants and eateries to satisfy every palate.
Families can enjoy captivating puppet shows and theatrical performances at this popular venue.
Every year on the second Saturday in August, the city comes alive with music, dance performances, and parades to honor the art of dance.
The theater is known for its talented actors and captivating performances, attracting theater enthusiasts from near and far.
These architectural gems are not only places of worship but also important cultural and historical landmarks.
Art enthusiasts can explore a diverse collection of paintings, sculptures, and other artistic creations at this prestigious museum.
The city offers a range of bars, clubs, and entertainment venues where locals and visitors can unwind and dance the night away.
Lined with trees and adorned with statues, the square serves as a meeting point for friends and a venue for outdoor concerts and events.
The city’s strategic location makes it an ideal starting point for exploring the natural wonders and unique cultures of the region.
With a vast collection of books, magazines, and digital resources, the library is a haven for bookworms and researchers.
Residents actively participate in various sports and athletic activities , fostering a healthy and active lifestyle.
Visitors can delve into the past and learn about the city’s origins and development through interactive displays and exhibits.
Here, locals can find fresh produce, delicious local delicacies, and a wide variety of goods and products.
Throughout the year, the city hosts a range of celebrations, including the Khabarovsk City Day and the Far Eastern Crafts Fair .
Nature enthusiasts can explore the lush greenery, beautiful flowers, and rare plant species within this peaceful garden.
Visitors to the city are greeted with open arms and a friendly smile, making them feel welcome and at home.
In conclusion, Khabarovsk is a fascinating city that offers a rich blend of history, culture, and natural beauty. From its picturesque waterfront and parks to its impressive landmarks and museums, there is much to explore and discover in this vibrant Russian city. Whether you’re interested in learning about its intriguing past, indulging in local cuisine, or simply enjoying the stunning surroundings , Khabarovsk has something to offer for everyone. So, why not plan a visit and immerse yourself in the charm and allure of this unique destination?
1. Where is Khabarovsk located?
Khabarovsk is located in the Far Eastern Federal District of Russia , near the border with China.
2. What is the best time to visit Khabarovsk?
The best time to visit Khabarovsk is during the summer months of June to August when the weather is pleasant and the city comes alive with festivals and outdoor activities.
3. How can I get to Khabarovsk?
Khabarovsk is well-connected by air, rail, and road. You can reach the city by taking a flight to Khabarovsk Novy Airport or by train from major Russian cities like Moscow and Vladivostok .
4. What are some must-visit attractions in Khabarovsk?
Some must-visit attractions in Khabarovsk include Khabarovsk Krai Museum, Lenin Square, Amur Cliff, and Muravyov-Amursky Street.
5. Is Khabarovsk a safe city for tourists?
Khabarovsk is generally considered safe for tourists. However, it is always advisable to take usual safety precautions and be aware of your surroundings, especially in crowded areas.
6. Are there any outdoor activities in Khabarovsk?
Yes, Khabarovsk offers plenty of outdoor activities such as boat trips along the Amur River, hiking in the nearby national parks, and enjoying picnics in the city’s beautiful parks and gardens.
7. Is English widely spoken in Khabarovsk?
English is not widely spoken in Khabarovsk, but you can usually find English-speaking staff at major tourist attractions, hotels, and some restaurants.
8. Are there any local specialties to try in Khabarovsk?
Yes, Khabarovsk is known for its delicious local cuisine. Some popular local specialties include ukha (fish soup), pelmeni (dumplings), and vareniki (stuffed dumplings).
9. Can I use credit cards in Khabarovsk?
Credit cards are widely accepted in major establishments such as hotels, restaurants, and shops. However, it is always advisable to carry some cash for smaller vendors and local markets.
10. Are there any day trips or excursions from Khabarovsk?
Yes, there are several day trips and excursions you can take from Khabarovsk, such as visiting the Sikhote-Alin Reserve, exploring the historic city of Blagoveshchensk, or taking a boat tour to the Bolshoi Ussuriysky Island.
Khabarovsk's captivating history, cultural diversity, and natural beauty make it a truly unique destination. From its vibrant sports scene, including the local football club , to the fascinating Negidal people who call the region home, there's always more to explore. Don't miss the chance to learn about Komsomolsk-na-Amure, another city in the Khabarovsk Krai with its own set of intriguing facts and stories.
Our commitment to delivering trustworthy and engaging content is at the heart of what we do. Each fact on our site is contributed by real users like you, bringing a wealth of diverse insights and information. To ensure the highest standards of accuracy and reliability, our dedicated editors meticulously review each submission. This process guarantees that the facts we share are not only fascinating but also credible. Trust in our commitment to quality and authenticity as you explore and learn with us.
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Photographs by William Brumfield
By rail, the city of Khabarovsk in the Russian Far East is a six-day, 5,300 mile journey from Moscow. Given the distance, it’s unsurprising that most visitors prefer to make the trip by air. The nine-hour flight has the added advantage of giving travelers a dramatic overview of the Amur River on the approach.
Indeed, Khabarovsk is a city of two great strategic rivers: the Amur, which flows eastward along the border with China; and the Ussuri, which flows northward some 560 miles from its origins in the southern Sikhote-Alin Mountains, not far from the Sea of Japan. The confluence of these two rivers near Khabarovsk, and the state boundaries they define have created one of the most important geopolitical zones in northern Asia.
Russian explorers attempted to gain control of the north bank of the Amur as early as the 1640s, but Russia was compelled to abandon the area by the Treaty of Nerchinsk, signed in 1689. Russian troops did not return to the region in substantial numbers until the mid-19 th century, under the leadership of Governor-General Nikolai Muravyov.
He was later given the title “Amursky” for his role in acquiring the territory for the Russian empire. In the Treaty of Aigun (1858), negotiated with the Qing Dynasty, China ceded to Russia territory north of the Amur and east of the Ussuri.
Khabarovsk arose as a consequence of the Treaty of Aigun. The first Russian settlement was established in May 1858 by a detachment of Siberian troops under the command of Captain Yakov Dyachenko. The post was initially named Khabarovka in homage to the renowned 17th-century Cossack leader Yerofei Khabarov, who explored the area.
The settlement rapidly expanded, and by 1864 it already had a formal plan for development on hilly terrain along the left bank of the Amur. A telegraph line to Vladivostok began operating in 1868, and a proper river port was completed in 1874. By the end of 1880, Khabarovsk had gained over 4,000 inhabitants and an official status as a town.
To get to Khabarovsk from Moscow of St. Petersburg take a regular flight. The trip takes approximately 7,5 hours.
Entrepreneurs looking to capitalize on the development of river trading routes were quick to settle in the young town, which offered considerable potential for commerce with China. Progress in transportation also led to increased government authority. In 1884, Khabarovsk became the administrative center of a vast area stretching from the Amur River to the Pacific.
The town’s status was further enhanced in late May 1891 when the heir to the throne, Tsarevich Nicholas Alexandrovich (subsequently Nicholas II) included Khabarovka on the itinerary of his nine-month-long world tour.
At the time of his visit, the town unveiled a heroic monument to Muravyov-Amursky by the noted sculptor Alexander Opekushin. Dismantled during the Soviet period, the monument has since been restored to a prominent position overlooking the river. In 1893, the town’s name was changed from Khabarovka to the more formal Khabarovsk.
Perhaps the greatest catalyst for the town’s early growth was the completion of a rail line from Vladivostok in 1897. Khabarovsk now had a direct link to a growing international port, even as it controlled interior river traffic over the extensive territory of the Amur River basin.
The town’s strategic location was not lost on military planners, who developed local machine factories to equip the armed forces of the Far East. At the same time, Khabarovsk, like the rest of Russia, experienced severe labor and military unrest in 1905-06 following the country’s defeat in the Russo-Japanese War.
With the return to stability under Prime Minister Pyotr Stolypin (1862-1911), Khabarovsk continued its rapid growth. During the three decades from 1884 to the beginning of World War I, the population increased more than 10-fold. And in 1908, Khabarovsk became the headquarters of the Amur River Flotilla, with responsibility for patrolling the long frontier with Manchuria.
In 1914, connections were improved with the Trans-Siberian Railway, which gave Khabarovsk access to distant Moscow and St. Petersburg. At that time, however, the rail route to the Russian Far East still went through Manchuria along the Chinese Eastern Railway.
The town was linked directly to eastern Siberia only in 1916, with the completion of a bridge across the Amur just to the northwest of Khabarovsk. Built in difficult conditions within three years, the bridge was one of the major achievements of Russian engineering.
To this day the attractive, often imposing architecture of central Khabarovsk reflects the prosperity of the town at the turn of the 20 th century. Using an eclectic mixture of neoclassicism and medieval Russian elements, architects designed enduring, well-built structures for housing, commerce and administration. A peculiar local feature was the use of high quality, unstuccoed red brick for the structure, with gray brick for decorative trim.
The most visible indicator of prosperity and rising consumer demand was the large department store. Firms such as Kunst and Albers, which had stores in several towns, and the Pyankov Brothers used architecture to create an impressive display for retail trade. The large Plyusnin building, subsequently converted to the Regional Library, contained one of the town’s many banks. The best of these buildings are on the main street, which is named after Muravyov-Amursky.
Other historic buildings display the style moderne that was fashionable at the beginning of the 20 th century, with traces of traditional Russian decoration. The best example is the former building of the city council, now carefully restored.
The Shrines of Kargopol: Preserving the art of the Russian North
The devastation of World War I occurred far from Khabarovsk, and the city actually grew with expanding military production. But the civil war following the Bolshevik Revolution caused major damage and disruption.
The last major battles of the Civil War occurred near Khabarovsk. Red partisan forces recaptured the town in early 1920, but they were suddenly attacked by the Japanese in April. Fierce fighting led to significant destruction in the central district. Instability continued until December 1921, when the city was retaken by a White army led by Viktorin Molchanov.
At the battle of Volochaevka in February 1922, Molchanov’s defenses were breached and Khabarovsk was retaken by Red forces, but not without further damage to the area, including partial demolition of the magnificent Amur River Bridge. Authority was vested in the Far Eastern Republic, a Communist ally that formally merged with the new Soviet state in November 1922.
During the Soviet period, the expansion of Khabarovsk accelerated thanks to its strategic military, industrial and administrative position. The pace of growth is reflected in modernist buildings designed by prominent Constructivist architects such as Ilya Golosov, who built the large complex for the House of Soviets in 1929-30.
In the 1930s the Gulag concentration camp empire expanded and prison labor was used in construction. A number of building projects in Khabarovsk were undertaken by the NKVD, which preferred a pompous neoclassical style. Relics of that time include the GlavDalStroi Building and the Commune House, all on Muravyov-Amursky Street.
World War II, like the first, occurred far from Khabarovsk, but the city played a major role in defending the Far East from a Japanese attack. And it served as headquarters during the brief Soviet-Japanese War in August 1945
Liavlia and Zaostrovye: Enduring traditions in the Arkhangelsk Region
Greatly expanded after the war, Khabarovsk maintained its momentum into the post-Soviet period. The city’s vitality is reflected in the improved appearance of Muravyov-Amursky Street, from a renovated Lenin Square to Cathedral Square, with its new Dormition Cathedral visible from the Amur. In 2004 the Transfiguration Cathedral was consecrated on Glory Square near the river.
With a population of just over 600,000, Khabarovsk has witnessed a building spree of contemporary apartment houses in colorful post-modernist forms. And the city’s good management was acknowledged in 2000 when it was chosen as the headquarters of the Far Eastern Federal District.
In the late summer of 2013 the Khabarovsk area experienced a record-setting flood of the Amur, but the main part of the city — on high ground — avoided the worst of the destruction. Among the city's promising economic developments is its pivotal role in the massive "Strength of Siberia" gas pipeline project, which will link the gas fields of Yakutia to the rapidly expanding Chinese market. With its well-maintained central district, Khabarovsk preserves its heritage as it looks to the future.
All rights reserved by Rossiyskaya Gazeta.
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Хабаровский край | |
Coordinates: 136°50′E / 54.800°N 136.833°E / 54.800; 136.833 | |
Country | |
Administrative center | |
Government | |
• Body | |
• | (acting) |
Area | |
• Total | 787,633 km (304,107 sq mi) |
• Rank | |
Population ( ) | |
• Total | 1,292,944 |
• Estimate | 1,328,302 |
• Rank | |
• Density | 1.6/km (4.3/sq mi) |
• | 83.4% |
• | 16.6% |
( ) | |
RU-KHA | |
27 | |
ID | 08000000 |
Official languages | |
Website |
Heavy industry, demographics, ethnic groups, settlements, sister relations, external links.
Being dominated by the Siberian High winter cold, the continental climates of the krai see extreme freezing for an area adjacent to the sea near the mid-latitudes, but also warm summers in the interior. The southern region lies mostly in the basin of the lower Amur River , with the mouth of the river located at Nikolaevsk-on-Amur draining into the Strait of Tartary , which separates Khabarovsk Krai from the island of Sakhalin . The north occupies a vast mountainous area along the coastline of the Sea of Okhotsk , a marginal sea of the Pacific Ocean . Khabarovsk Krai is bordered by Magadan Oblast to the north; Amur Oblast , Jewish Autonomous Oblast , and the Sakha Republic to the west; Primorsky Krai to the south; and Sakhalin Oblast to the east.
The population consists of mostly ethnic Russians , but indigenous people of the area are numerous, such as the Tungusic peoples ( Evenks , Negidals , Ulchs , Nanai , Oroch , Udege ), Amur Nivkhs , and Ainu . [10]
Khabarovsk Krai shares its borders with Magadan Oblast in the north; with the Sakha Republic and Amur Oblast in the west; with the Jewish Autonomous Oblast , China ( Heilongjiang ), and Primorsky Krai in the south; and is limited by the Sea of Okhotsk in the east. In terms of area, it is the fourth-largest federal subject within Russia. Major islands include the Shantar Islands .
Taiga and tundra in the north, swampy forest in the central depression, and deciduous forest in the south are the natural vegetation in the area. The main rivers are the Amur , Amgun , Uda , and Tugur , among others. There are also lakes such as Bokon , Bolon , Chukchagir , Evoron , Kizi , Khummi , Orel , and Udyl , among others. [11]
Khabarovsk Krai has a severely continental climate with its northern areas being subarctic with stronger maritime summer moderation in the north. In its southerly areas, especially inland, annual swings are extremely strong, with Khabarovsk itself having hot, wet, and humid summers which rapidly transform into severely cold and long winters, where temperatures hardly ever go above freezing. This is because of the influence of the East Asian monsoon in summer and the bitterly cold Siberian High in winter. The second-largest city of Komsomolsk-on-Amur has even more violent temperature swings than Khabarovsk, with winter average lows below −30 °C (−22 °F) , but in spite of this, avoiding being subarctic because of the significant heat in summer.
The main mountain ranges in the region are the Bureya Range , the Badzhal Range (highest point 2,221 metres (7,287 ft) high, the Gora Ulun ), the Yam-Alin , the Dusse-Alin , the Sikhote-Alin , the Dzhugdzhur Mountains , the Kondyor Massif , as well as a small section of the Suntar-Khayata Range , the Yudoma-Maya Highlands , and the Sette-Daban in the western border regions. The highest point is 2,933 metres (9,623 ft) high, Berill Mountain . [12] [13]
There are a number of peninsulas along the krai's extensive coast, the main ones being (north to south) the Lisyansky Peninsula , Nurki Peninsula , Tugurskiy Peninsula , and the Tokhareu Peninsula .
The main islands of Khabarovsk Krai (north to south) are Malminskiye Island , the Shantar Islands , Menshikov Island , Reyneke Island (Sea of Okhotsk) , Chkalov Island , Baydukov Island , and the Chastye Islands . The island of Sakhalin (Russia's largest) is administered separately as Sakhalin Oblast , along with the Kuril Islands .
The charts below detail climate averages from various locations in the krai. Khabarovsk is set near the Chinese border at a lower latitude far inland, while Komsomolsk-on-Amur being further downstream on the Amur river at a higher latitude. Sovetskaya Gavan and Okhotsk are coastal settlements in the deep south and far north, respectively.
Climate data for (1991–2020, extremes 1878–2023) | |||||||||||||
---|---|---|---|---|---|---|---|---|---|---|---|---|---|
Month | Jan | Feb | Mar | Apr | May | Jun | Jul | Aug | Sep | Oct | Nov | Dec | Year |
Record high °C (°F) | 0.6 (33.1) | 6.3 (43.3) | 17.0 (62.6) | 28.6 (83.5) | 31.5 (88.7) | 36.4 (97.5) | 35.7 (96.3) | 35.6 (96.1) | 29.8 (85.6) | 26.4 (79.5) | 15.5 (59.9) | 6.6 (43.9) | 36.4 (97.5) |
Mean daily maximum °C (°F) | −14.9 (5.2) | −9.9 (14.2) | −1.0 (30.2) | 10.5 (50.9) | 19.2 (66.6) | 23.8 (74.8) | 26.8 (80.2) | 24.9 (76.8) | 19.7 (67.5) | 10.6 (51.1) | −2.8 (27.0) | −13.6 (7.5) | 7.8 (46.0) |
Daily mean °C (°F) | −19.2 (−2.6) | −14.9 (5.2) | −5.9 (21.4) | 4.8 (40.6) | 12.9 (55.2) | 18.0 (64.4) | 21.4 (70.5) | 19.9 (67.8) | 14.1 (57.4) | 5.4 (41.7) | −6.9 (19.6) | −17.4 (0.7) | 2.7 (36.9) |
Mean daily minimum °C (°F) | −23.1 (−9.6) | −19.6 (−3.3) | −10.7 (12.7) | −0.1 (31.8) | 7.3 (45.1) | 12.8 (55.0) | 16.8 (62.2) | 15.7 (60.3) | 9.4 (48.9) | 1.0 (33.8) | −10.4 (13.3) | −20.9 (−5.6) | −1.8 (28.8) |
Record low °C (°F) | −40.0 (−40.0) | −35.1 (−31.2) | −28.9 (−20.0) | −15.1 (4.8) | −3.1 (26.4) | 2.2 (36.0) | 6.8 (44.2) | 4.9 (40.8) | −3.3 (26.1) | −15.6 (3.9) | −27.7 (−17.9) | −38.1 (−36.6) | −40.0 (−40.0) |
Average mm (inches) | 13 (0.5) | 12 (0.5) | 22 (0.9) | 37 (1.5) | 70 (2.8) | 84 (3.3) | 137 (5.4) | 143 (5.6) | 85 (3.3) | 48 (1.9) | 26 (1.0) | 19 (0.7) | 696 (27.4) |
Average extreme snow depth cm (inches) | 14 (5.5) | 16 (6.3) | 12 (4.7) | 1 (0.4) | 0 (0) | 0 (0) | 0 (0) | 0 (0) | 0 (0) | 1 (0.4) | 5 (2.0) | 10 (3.9) | 16 (6.3) |
Average rainy days | 0 | 0 | 1 | 10 | 16 | 15 | 15 | 17 | 15 | 11 | 2 | 0 | 102 |
Average snowy days | 14 | 11 | 11 | 6 | 1 | 0 | 0 | 0 | 0.1 | 4 | 12 | 14 | 73 |
Average (%) | 75 | 72 | 68 | 63 | 65 | 74 | 79 | 83 | 78 | 67 | 69 | 73 | 72 |
Mean monthly | 147 | 181 | 231 | 213 | 242 | 262 | 248 | 217 | 212 | 189 | 159 | 145 | 2,446 |
Source 1: Pogoda.ru.net | |||||||||||||
Source 2: NOAA (sun, 1961–1990) |
Climate data for | |||||||||||||
---|---|---|---|---|---|---|---|---|---|---|---|---|---|
Month | Jan | Feb | Mar | Apr | May | Jun | Jul | Aug | Sep | Oct | Nov | Dec | Year |
Record high °C (°F) | 0.7 (33.3) | 0.0 (32.0) | 13.6 (56.5) | 23.9 (75.0) | 31.0 (87.8) | 33.2 (91.8) | 36.2 (97.2) | 38.0 (100.4) | 30.0 (86.0) | 20.5 (68.9) | 8.3 (46.9) | 1.0 (33.8) | 38.0 (100.4) |
Mean daily maximum °C (°F) | −19.6 (−3.3) | −13.9 (7.0) | −4.0 (24.8) | 7.5 (45.5) | 16.1 (61.0) | 22.8 (73.0) | 25.1 (77.2) | 23.4 (74.1) | 17.1 (62.8) | 7.4 (45.3) | −6.4 (20.5) | −17.2 (1.0) | 4.6 (40.3) |
Daily mean °C (°F) | −24.7 (−12.5) | −19.8 (−3.6) | −9.5 (14.9) | 2.3 (36.1) | 10.4 (50.7) | 17.3 (63.1) | 20.3 (68.5) | 18.5 (65.3) | 11.9 (53.4) | 2.5 (36.5) | −10.5 (13.1) | −21.8 (−7.2) | −0.6 (30.9) |
Mean daily minimum °C (°F) | −30.8 (−23.4) | −27.2 (−17.0) | −17.1 (1.2) | −3.4 (25.9) | 3.7 (38.7) | 10.8 (51.4) | 15.2 (59.4) | 13.5 (56.3) | 6.4 (43.5) | −2.9 (26.8) | −16.1 (3.0) | −27.4 (−17.3) | −6.6 (20.1) |
Record low °C (°F) | −47.0 (−52.6) | −42.0 (−43.6) | −33.9 (−29.0) | −20.8 (−5.4) | −7.5 (18.5) | −2.2 (28.0) | 0.0 (32.0) | −8.9 (16.0) | −6.0 (21.2) | −22.0 (−7.6) | −34.0 (−29.2) | −42.0 (−43.6) | −47.0 (−52.6) |
Average mm (inches) | 30 (1.2) | 19 (0.7) | 30 (1.2) | 43 (1.7) | 63 (2.5) | 65 (2.6) | 95 (3.7) | 110 (4.3) | 74 (2.9) | 62 (2.4) | 49 (1.9) | 32 (1.3) | 672 (26.4) |
Average precipitation days | 14 | 12 | 13 | 15 | 15 | 13 | 15 | 14 | 14 | 13 | 16 | 15 | 169 |
Average rainy days | 0 | 0 | 1 | 7 | 14 | 13 | 15 | 14 | 14 | 8 | 1 | 0 | 87 |
Average snowy days | 14 | 12 | 13 | 11 | 3 | 0 | 0 | 0 | 0 | 8 | 15 | 15 | 91 |
Source 1: climatebase.ru | |||||||||||||
Source 2: Weatherbase |
Climate data for (1914–2012) | |||||||||||||
---|---|---|---|---|---|---|---|---|---|---|---|---|---|
Month | Jan | Feb | Mar | Apr | May | Jun | Jul | Aug | Sep | Oct | Nov | Dec | Year |
Record high °C (°F) | 2.6 (36.7) | 12.2 (54.0) | 18.9 (66.0) | 25.1 (77.2) | 31.8 (89.2) | 35.1 (95.2) | 34.2 (93.6) | 35.8 (96.4) | 30.2 (86.4) | 26.8 (80.2) | 16.5 (61.7) | 9.4 (48.9) | 35.8 (96.4) |
Mean daily maximum °C (°F) | −11.4 (11.5) | −8.3 (17.1) | −1.8 (28.8) | 5.6 (42.1) | 11.6 (52.9) | 16.8 (62.2) | 20.5 (68.9) | 21.9 (71.4) | 18.2 (64.8) | 10.9 (51.6) | 0.0 (32.0) | −8.7 (16.3) | 6.3 (43.3) |
Daily mean °C (°F) | −16.8 (1.8) | −14.2 (6.4) | −7.4 (18.7) | 1.1 (34.0) | 6.6 (43.9) | 11.5 (52.7) | 15.6 (60.1) | 17.4 (63.3) | 13.3 (55.9) | 6.0 (42.8) | −4.7 (23.5) | −13.5 (7.7) | 1.3 (34.3) |
Mean daily minimum °C (°F) | −22.2 (−8.0) | −20.1 (−4.2) | −12.9 (8.8) | −3.5 (25.7) | 1.5 (34.7) | 6.2 (43.2) | 10.7 (51.3) | 12.9 (55.2) | 8.4 (47.1) | 1.0 (33.8) | −9.3 (15.3) | −18.3 (−0.9) | −3.8 (25.2) |
Record low °C (°F) | −40.0 (−40.0) | −38.6 (−37.5) | −30.3 (−22.5) | −26.4 (−15.5) | −9.5 (14.9) | −3.0 (26.6) | 2.4 (36.3) | 4.0 (39.2) | −1.7 (28.9) | −14.7 (5.5) | −31.3 (−24.3) | −38.4 (−37.1) | −40.0 (−40.0) |
Average mm (inches) | 19.9 (0.78) | 20.7 (0.81) | 42.9 (1.69) | 47.5 (1.87) | 73.9 (2.91) | 70.1 (2.76) | 82.1 (3.23) | 109.6 (4.31) | 117.2 (4.61) | 87.7 (3.45) | 43.4 (1.71) | 32.7 (1.29) | 747.7 (29.42) |
Average precipitation days | 6.8 | 7.0 | 9.6 | 10.3 | 13.2 | 12.9 | 13.4 | 14.7 | 13.1 | 9.2 | 6.1 | 6.6 | 122.9 |
Source: |
Climate data for (1991−2020 normals, extremes 1891–present) | |||||||||||||
---|---|---|---|---|---|---|---|---|---|---|---|---|---|
Month | Jan | Feb | Mar | Apr | May | Jun | Jul | Aug | Sep | Oct | Nov | Dec | Year |
Record high °C (°F) | 5.5 (41.9) | 2.0 (35.6) | 6.4 (43.5) | 16.0 (60.8) | 26.2 (79.2) | 31.3 (88.3) | 31.0 (87.8) | 32.1 (89.8) | 24.8 (76.6) | 15.7 (60.3) | 6.2 (43.2) | 2.8 (37.0) | 32.1 (89.8) |
Mean daily maximum °C (°F) | −16.8 (1.8) | −14.2 (6.4) | −6.3 (20.7) | 0.4 (32.7) | 6.2 (43.2) | 11.4 (52.5) | 15.7 (60.3) | 17.1 (62.8) | 12.9 (55.2) | 2.7 (36.9) | −9.7 (14.5) | −16.4 (2.5) | 0.3 (32.5) |
Daily mean °C (°F) | −19.9 (−3.8) | −18.5 (−1.3) | −12.1 (10.2) | −3.8 (25.2) | 2.6 (36.7) | 8.1 (46.6) | 12.9 (55.2) | 13.7 (56.7) | 8.9 (48.0) | −1.2 (29.8) | −12.7 (9.1) | −19.0 (−2.2) | −3.4 (25.9) |
Mean daily minimum °C (°F) | −22.7 (−8.9) | −22.2 (−8.0) | −17.8 (0.0) | −8.2 (17.2) | −0.2 (31.6) | 5.7 (42.3) | 10.6 (51.1) | 10.6 (51.1) | 4.9 (40.8) | −4.6 (23.7) | −15.3 (4.5) | −21.4 (−6.5) | −6.7 (19.9) |
Record low °C (°F) | −41.3 (−42.3) | −45.7 (−50.3) | −36.9 (−34.4) | −29.2 (−20.6) | −16.0 (3.2) | −2.6 (27.3) | 1.7 (35.1) | −0.1 (31.8) | −6.6 (20.1) | −27.5 (−17.5) | −37.4 (−35.3) | −37.7 (−35.9) | −45.7 (−50.3) |
Average mm (inches) | 15 (0.6) | 7 (0.3) | 16 (0.6) | 24 (0.9) | 40 (1.6) | 55 (2.2) | 85 (3.3) | 94 (3.7) | 92 (3.6) | 66 (2.6) | 32 (1.3) | 14 (0.6) | 540 (21.3) |
Average rainy days | 0.1 | 0.2 | 0.3 | 2 | 11 | 16 | 18 | 15 | 16 | 7 | 1 | 0.2 | 87 |
Average snowy days | 9 | 9 | 11 | 13 | 10 | 0.4 | 0 | 0 | 0.3 | 9 | 11 | 8 | 81 |
Average (%) | 63 | 63 | 68 | 77 | 84 | 88 | 89 | 86 | 80 | 70 | 66 | 63 | 75 |
Mean monthly | 86 | 147 | 241 | 230 | 195 | 200 | 179 | 182 | 172 | 157 | 107 | 54 | 1,950 |
Source 1: Pogoda.ru.net | |||||||||||||
Source 2: (sun 1961–1990) |
According to various Chinese and Korean records, the southern part of Khabarovsk Krai was originally occupied by one of the five semi-nomadic Shiwei , the Bo Shiwei tribes, and the Black Water Mohe tribes living, respectively, on the west and the east of the Bureya and the Lesser Khingan ranges.
In 1643, Vassili Poyarkov 's boats descended the Amur , returning to Yakutsk by the Sea of Okhotsk and the Aldan River , and in 1649–1650, Yerofey Khabarov occupied the banks of the Amur. The resistance of the Chinese, however, obliged the Cossacks to quit their forts, and by the Treaty of Nerchinsk (1689), Russia abandoned its advance into the basin of the river.
Although the Russians were thus deprived of the right to navigate the Amur River, the territorial claim over the lower courses of the river was not settled in the Treaty of Nerchinsk of 1689. The area between the Uda River and the Greater Khingan mountain range (i.e. most of Lower Amuria) was left undemarcated and the Sino-Russian border was allowed to fluctuate. [20] [21]
Later in the nineteenth century, Nikolay Muravyov conducted an aggressive policy with China by claiming that the lower reaches of the Amur River belonged to Russia . In 1852, a Russian military expedition under Muravyov explored the Amur, and by 1857, a chain of Russian Cossacks and peasants had been settled along the whole course of the river. In 1858, in the Treaty of Aigun , China recognized the Amur River downstream as far as the Ussuri River as the boundary between Russia and the Qing Empire, and granted Russia free access to the Pacific Ocean. [22] The Sino-Russian border was later further delineated in the Treaty of Peking of 1860 when the Ussuri Territory (the Maritime Territory ), which was previously a joint possession, became Russian. [23]
Khabarovsk Krai was established on 20 October 1938, when the Far Eastern Krai was split into the Khabarovsk and Primorsky Krais . [24] Kamchatka Oblast , which was originally subordinated to the Far Eastern Krai, fell under the Jurisdiction of Khabarovsk Krai, along with its two National Okrugs, Chukotka and Koryak . In 1947, the northern part of Sakhalin was removed from the Krai to join the southern part and form Sakhalin Oblast . In 1948, parts of its southwestern territories were removed from the Krai to form Amur Oblast . In 1953, Magadan Oblast was established from the northern parts of the Krai and was given jurisdiction over Chukotka National Okrug, which was originally under the jurisdiction of Kamchatka oblast. In 1956, Kamchatka Oblast became its own region and took Koryak National Okrug with it. The Krai took its modern form in 1991, just before the USSR's collapse when the Jewish Autonomous Oblast was created within its territory. On 24 April 1996, Khabarovsk signed a power-sharing agreement with the federal government, granting it autonomy. [25] This agreement would be abolished on 12 August 2002. [26]
During the Soviet period, the high authority in the oblast was shared between three persons: The first secretary of the Khabarovsk CPSU Committee (who, in reality, had the biggest authority), the chairman of the oblast Soviet (legislative power), and the Chairman of the oblast Executive Committee (executive power). Since 1991, CPSU lost all the power, and the head of the Oblast administration, and eventually the governor, was appointed/elected alongside elected regional parliament .
The Charter of Khabarovsk Krai is the fundamental law of the krai. The Legislative Duma of Khabarovsk Krai is the regional standing legislative (representative) body. The Legislative Duma exercises its authority by passing laws, resolutions, and other legal acts and by supervising the implementation and observance of the laws and other legal acts passed by it. The highest executive body is the Krai Government, which includes territorial executive bodies, such as district administrations, committees, and commissions that facilitate development and run the day to day matters of the province. The Krai Administration supports the activities of the Governor , who is the highest official and acts as guarantor of the observance of the Charter in accordance with the Constitution of Russia .
On 9 July 2020, the governor of the region, Sergei Furgal , was arrested and flown to Moscow. The 2020 Khabarovsk Krai protests began on 11 July 2020, in support of Furgal. [27]
Khabarovsk Krai is the most industrialized territory of the Far East of Russia, producing 30% of the total industrial products in the Far Eastern Economic Region.
The machine construction industry consists primarily of a highly developed military–industrial complex of large-scale aircraft- and shipbuilding enterprises. [28] The Komsomolsk-on-Amur Aircraft Production Association is currently among the krai's most successful enterprises, and for years has been the largest taxpayer of the territory. [28] Other major industries include timber-working and fishing , along with metallurgy in the main cities. Komsomolsk-on-Amur is the iron and steel centre of the Far East; a pipeline from northern Sakhalin supplies the petroleum-refining industry in the city of Khabarovsk . In the Amur basin, there is also some cultivation of wheat and soybeans . The administrative centre , Khabarovsk, is at the junction of the Amur River and the Trans-Siberian Railway .
The region's mineral resources are relatively underdeveloped. Khabarovsk Krai contains large gold mining operations (Highland Gold, Polus Gold), a major but low-grade copper deposit being explored by IG Integro Group , and a world-class tin district which was a major contributor to the Soviet industrial complex and is currently being revitalised by Far Eastern Tin (Festivalnoye mine) and by Sable Tin Resources Archived March 13, 2017, at the Wayback Machine , which is developing the Sable Tin Deposit (Sobolinoye) , a large high-grade deposit, 25 km from Solnechny town.
Year | ||
---|---|---|
1926 | 184,700 | — |
1939 | 657,400 | +255.9% |
1959 | 979,679 | +49.0% |
1970 | 1,173,458 | +19.8% |
1979 | 1,369,277 | +16.7% |
1989 | 1,597,373 | +16.7% |
2002 | 1,436,570 | −10.1% |
2010 | 1,343,869 | −6.5% |
2021 | 1,292,944 | −3.8% |
Source: Census data |
Population : 1,292,944 ( 2021 Census ) ; [29] 1,343,869 ( 2010 Russian census ) ; [9] 1,436,570 ( 2002 Census ) ; [30] 1,824,506 ( 1989 Soviet census ) . [31]
Ethnicity | Population | Percentage |
---|---|---|
1,047,221 | 92.9% | |
10,813 | 1.0% | |
7,170 | 0.6% | |
4,332 | 0.4% | |
3,740 | 0.3% | |
3,709 | 0.3% | |
Other Ethnicities | 50,780 | 3.9% |
Ethnicity not stated | 165,179 | – |
Vital statistics for 2022: [33] [34]
Total fertility rate (2022): [35] 1.50 children per woman
Life expectancy (2021): [36] Total — 67.85 years (male — 62.91, female — 72.94)
| |||||||||
---|---|---|---|---|---|---|---|---|---|
Rank | Pop. | ||||||||
| 1 | 577,441 | | ||||||
2 | 263,906 | ||||||||
3 | 42,970 | ||||||||
4 | 22,752 | ||||||||
5 | 27,712 | ||||||||
6 | 17,154 | ||||||||
7 | 17,001 | ||||||||
8 | 14,555 | ||||||||
9 | 13,306 | ||||||||
10 | 13,048 |
Religion in Krai Oblast as of 2012 (Sreda Arena Atlas) | ||||
---|---|---|---|---|
26.2% | ||||
Other | 1.3% | |||
0.5% | ||||
Other | 3.7% | |||
1.1% | ||||
and other native faiths | 0.5% | |||
27.9% | ||||
and | 23.1% | |||
Other and undeclared | 15.7% |
According to a 2012 survey, [37] 26.2% of the population of Khabarovsk Krai adheres to the Russian Orthodox Church , 4% are unaffiliated generic Christians , 1% adhere to other Orthodox churches or are believers in Orthodox Christianity who do not belong to any church, while 1% are adherents of Islam . In addition, 28% of the population declared to be "spiritual but not religious", 23% are atheist , and 16.8% follow other religions or did not give an answer to the question. [37]
There are the following institutions of higher education in Khabarovsk Krai. [39] [40]
The city was a host to the 1981 Bandy World Championship as well as to the 2015 Bandy World Championship . For the 2015 games, twenty-one teams originally were expected, which would have been four more than the record-making seventeen from the 2014 tournament , but eventually, only sixteen teams came. The A Division of the 2018 Bandy World Championship was again to be played in Khabarovsk. [42]
Amur Oblast is a federal subject of Russia, located on the banks of the Amur and Zeya rivers in the Russian Far East. Amur Oblast borders Heilongjiang province of the People's Republic of China (PRC) to the south.
Okha is a town and the administrative center of Okhinsky District of Sakhalin Oblast, Russia. Population: 23,008 (2010 Russian census) ; 27,963 (2002 Census) ; 36,104 (1989 Soviet census) .
Sakhalin Oblast is a federal subject of Russia comprising the island of Sakhalin and the Kuril Islands in the Russian Far East. The oblast has an area of 87,100 square kilometers (33,600 sq mi). Its administrative center and largest city is Yuzhno-Sakhalinsk. As of the 2021 Census, the oblast has a population of roughly 500,000.
Komsomolsk-on-Amur is a city in Khabarovsk Krai, Russia, located on the west bank of the Amur River in the Russian Far East. It is located on the Baikal-Amur Mainline, 356 kilometers (221 mi) northeast of Khabarovsk. Population: 238,505 (2021 Census) ; 263,906 (2010 Russian census) ; 281,035 (2002 Census) ; 315,325 (1989 Soviet census) .
Nikolayevsk-on-Amur is a town in Khabarovsk Krai, Russia located on the Amur River close to its liman in the Pacific Ocean. Population: 22,752 (2010 Russian census) ; 28,492 (2002 Census) ; 36,296 (1989 Soviet census) .
Sovetskaya Gavan is a town in Khabarovsk Krai, Russia, and a port on the Strait of Tartary which connects the Sea of Okhotsk in the north with the Sea of Japan in the south. Population: 27,712 (2010 Russian census) ; 30,480 (2002 Census) ; 34,915 (1989 Soviet census) .
Dolinsk is a town and the administrative center of Dolinsky District of Sakhalin Oblast, Russia, located in the southeast of the Sakhalin Island in the valley of the Naiba River and its tributaries, about 45 kilometers (28 mi) north of Yuzhno-Sakhalinsk and 10 kilometers (6.2 mi) from the coast of the Terpeniye Bay of the Sea of Okhotsk. Population: 12,200 (2010 Russian census) ; 12,555 (2002 Census) ; 15,653 (1989 Soviet census) .
Kamchatka Krai is a federal subject of Russia, situated in the Russian Far East. It is administratively part of the Far Eastern Federal District. Its administrative center and largest city is Petropavlovsk-Kamchatsky, home to over half of its population of 291,705.
Amursk is a town in Khabarovsk Krai, Russia, located on the left bank of the Amur River 45 kilometers (28 mi) south of Komsomolsk-on-Amur. Population: 42,970 (2010 Russian census) ; 47,759 (2002 Census) ; 58,395 (1989 Soviet census) .
Lesozavodsk is a town in Primorsky Krai, Russia, located on the Ussuri River, 10 kilometers (6.2 mi) from the Sino–Russian border and about 300 kilometers (190 mi) north of Vladivostok, the administrative center of the krai. Population: 37,034 (2010 Russian census) ; 42,185 (2002 Census) ; 44,065 (1989 Soviet census) ; 37,000 (1972).
Poronaysk is a town and the administrative center of Poronaysky District of Sakhalin Oblast, Russia, located on the Poronay River 288 kilometers (179 mi) north of Yuzhno-Sakhalinsk. Population: 16,120 (2010 Russian census) ; 17,954 (2002 Census) ; 25,971 (1989 Soviet census) .
Ayano-Maysky District is an administrative and municipal district (raion), one of the seventeen in Khabarovsk Krai, Russia. It is located in the north of the krai. The area of the district is 167,200 square kilometers (64,600 sq mi). Its administrative center is the rural locality of Ayan. Population: 2,292 (2010 Russian census) ; 3,271 (2002 Census) ; 4,802 (1989 Soviet census) . The population of Ayan accounts for 42.2% of the district's total population.
Tomari is a coastal town and the administrative center of Tomarinsky District in Sakhalin Oblast, Russia, located on the western coast of the Sakhalin Island, 167 kilometers (104 mi) northwest of Yuzhno-Sakhalinsk, the administrative center of the oblast. As of the 2010 Census, its population was 4,541.
Novy Urgal is an urban locality in Verkhnebureinsky District of Khabarovsk Krai, Russia, located in the valley of the Bureya River, close to its confluence with the Urgal River, about 340 kilometers (210 mi) northwest of the krai's administrative center of Khabarovsk and 28 kilometers (17 mi) west of the district's administrative center of Chegdomyn. Population: 6,803 (2010 Russian census) ; 7,274 (2002 Census) ; 9,126 (1989 Soviet census) .
Nogliki is an urban locality and the administrative center of Nogliksky District of Sakhalin Oblast, Russia, located near the eastern coast of Sakhalin Island, about 6 kilometers (3.7 mi) inland from the Sea of Okhotsk shoreline and about 600 kilometers (370 mi) north of Yuzhno-Sakhalinsk. Population: 10,231 (2010 Russian census) ; 10,729 (2002 Census) ; 11,546 (1989 Soviet census) .
Komsomolsky District is an administrative and municipal district (raion), one of the seventeen in Khabarovsk Krai, Russia. It is located in the southern central part of the krai. The area of the district is 25,167 square kilometers (9,717 sq mi). Its administrative center is the city of Komsomolsk-on-Amur. Population: 29,072 (2010 Russian census) ; 31,563 (2002 Census) ; 33,649 (1989 Soviet census) .
Nikolayevsky District is an administrative and municipal district (raion), one of the seventeen in Khabarovsk Krai, Russia. It is located in the east of the krai. The area of the district is 17,188 square kilometers (6,636 sq mi). Its administrative center is the town of Nikolayevsk-on-Amur. Population: 9,942 (2010 Russian census) ; 13,850 (2002 Census) ; 19,683 (1989 Soviet census) .
Okhotsky District is an administrative and municipal district (raion), one of the seventeen in Khabarovsk Krai, Russia. It is located in the north of the krai. The area of the district is 158,517.8 square kilometers (61,204.1 sq mi). Its administrative center is the urban locality of Okhotsk. Population: 8,197 (2010 Russian census) ; 12,017 (2002 Census) ; 19,183 (1989 Soviet census) . The population of Okhotsk accounts for 51.4% of the district's total population.
Tuguro-Chumikansky District is an administrative and municipal district (raion), one of the seventeen in Khabarovsk Krai, Russia. It is located in the center of the krai. The area of the district is 96,069 square kilometers (37,092 sq mi). Its administrative center is the rural locality of Chumikan. Population: 2,255 (2010 Russian census) ; 2,860 (2002 Census) ; 3,610 (1989 Soviet census) . The population of Chumikan accounts for 47.0% of the district's total population.
Selikhino is a rural locality in Komsomolsky District of Khabarovsk Krai, Russia. Population: 4,255 (2010 Russian census) ; 4,865 (2002 Census) .
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Daylight saving time (dst) changes in khabarovsk, sunrise and sunset time for khabarovsk, time and time zones.
The length of a solar day is determined by the time that it takes for the Earth to complete a full rotation around its axis and equals 24 hours. The Earth’s rotation on its axis leads to change between day and night. Another consequence of this rotation is the fact that while moving by 15° from West to East local solar time increases by an hour.
In everyday life people use official local time which almost always differs from solar time. All of the Earth’s surface is divided into time zones. All places within the same time zone observe the same time. Time zone boundaries usually follow country or administrative borders. Time difference between adjacent time zones normally equals one hour, though sometimes time in neighbouring time zones may differ by two or more hours. There are also cases when adjacent time zone difference equals 30 or 45 minutes.
For most countries the entire country’s territory lies within the same time zone. Countries whose territory stretches from West to East by a significant distance, such as Russia , USA , Canada , Brazil and some others, are usually divided into a few time zones. One notable exception is China where Beijing time serves as the official time all over the country.
Coordinated Universal Time or UTC is the reference point to determine time zone offsets. UTC corresponds to mean solar time on the Prime or Greenwich Meridian (0° longitude). Time zone offsets from UTC range from UTC-12:00 to UTC+14:00.
Almost all countries in Europe and North America as well as many other countries observe Daylight Saving Time (DST) and put their clocks an hour forward in the spring and an hour back in the autumn. In these countries time zone offsets from UTC change twice a year. Most countries do not observe DST though.
COMMENTS
Newton's Second Law and Problem Solving (PDF) The Curriculum Corner contains a complete ready-to-use curriculum for the high school physics classroom. This collection of pages comprise worksheets in PDF format that developmentally target key concepts and mathematics commonly covered in a high school physics curriculum.
The coefficient of friction between the crate and the floor is 0.750. Determine the acceleration of the crate. Given: m = 28.6 kg μ = 0.750 Fapp = 302 N Find a Begin by finding Fgrav: Fgrav = m•g = (28.6 kg)•(9.8 N/kg) = 280.28 N The Fnorm and Fgrav balance; so Fnorm is also 280.28 N. This value can be used to determine Ffrict: Ffrict = μ ...
Problem 22: Brandon is the catcher for the Varsity baseball team. He exerts a forward force on the .145-kg baseball to bring it to rest from a speed of 38.2 m/s. During the process, his hand recoils a distance of 0.135 m. Determine the acceleration of the ball and the force which is applied to it by Brandon.
The Physics Classroom, 2020. Page 1 Newton's Laws. A basketball star exerts a force of 3225 N (average value) upon the gym floor in order to accelerate his 76.5-kg body upward. (a) Determine the acceleration of the player. (b) Determine the final speed of the player if the force endures for a time of 0.150 seconds.
There are numerous resources at The Physics Classroom website that serve as very complementary supports for the Solve It! (with Newton's Second Law) Concept Builder. ... Each chapter of the Calculator Pad includes a problem set with as many as 35 problems. Each problem is accompanied by an answer that is revealed by tapping a button. And there ...
Answers: See answers and explanations below. 16. C - As the helicopter blades spin and push air in one direction, the air pushes the blades in the opposite direction; the result is that the helicopter can begin to rotate about the axis of the blade. To counteract this rotation, a second set of blades is required. 17.
ΣF = 3 N + 4 N + 2 N = 9 Newton, rightward. The equation of Newton's second law : ΣF = m a. a = ΣF / m. a = acceleration, ΣF = net force, m = mass. Based on the above formula, the acceleration (a) is directly proportional to the net force (ΣF) and inversely proportional to mass (m).
Khanmigo is now free for all US educators! Plan lessons, develop exit tickets, and so much more with our AI teaching assistant.
The second step is to solve for the unknown, in this case using Newton's second law. Finally, we check our answer to ensure it is reasonable. These techniques for integrated concept problems will be useful in applications of physics outside of a physics course, such as in your profession, in other science disciplines, and in everyday life.
Newton's second law describes the affect of net force and mass upon the acceleration of an object. Often expressed as the equation a = Fnet/m (or rearranged to Fnet=m*a), the equation is probably the most important equation in all of Mechanics. It is used to predict how an object will accelerated (magnitude and direction) in the presence of an unbalanced force.
Correct answer: Yes. Explanation: Newton's second law states that the net force, or the vector sum of all the forces acting on an object, equals the mass times the acceleration. So, it is possible to have forces act on an object without acceleration if the forces are oriented such that they vector sum to zero.
The relationship is stated by Newton's second law of motion, Force=Mass x Acceleration. -or-. F=ma. where F is the force, m is the mass, and a is the acceleration. The units are Newtons (N) for force, kilograms (kg) for mass, and meters per second squared (m/s2) for acceleration. The other forms of the equation can be used to solve for mass or ...
The Curriculum Corner contains a complete ready-to-use curriculum for the high school physics classroom. This collection of pages comprise worksheets in PDF format that developmentally target key concepts and mathematics commonly covered in a high school physics curriculum.
NEWTON'S LAWS WORKSHEET - KEY. I. NEWTON'S FIRST LAW OF MOTION. 1. Newton's first law of mo2on is also known as the LAW OF INERTIA. 2. Newton's first law says that. a. an object that IS NOT MOVING, or is at AT REST will stay at AT REST AND. b. an object that IS MOVING will keep moving with constant VELOCITY which means at the same SPEED ...
This worksheet will prepare students to solve simple problems including force calculations using Newton's 2nd Law of Motion. By completing this activity, 8th and 9th grade science students will learn how to calculate force problems. Included skills: Simple problems to introduce force calculations using Newton's 2nd law of motion.
The acceleration value can be determined using Newton's second law of motion. a = F net / m = (2.43x10 5 N) / (6.32x10 4 kg) = 3.84 m/s/s, left This acceleration value can be combined with other kinematic variables (vi = 94.3 km/hr = 26.2 m/s; t = 3.40 s) in order to determine the distance the train travels in 3.4 seconds.
36 Facts About Khabarovsk. Khabarovsk is a fascinating city filled with rich history, vibrant culture, and breathtaking natural landscapes. Located in the Far East region of Russia, Khabarovsk is the administrative center of Khabarovsk Krai. With its intriguing past and flourishing present, it is a city that captures the imagination of both ...
Dec 06 2014. William Brumfield. special to RBTH. Follow Russia Beyond on Facebook. This city in Russia's Far East is closer to the major cities of China than to Moscow. Photographs by William ...
Geography. Khabarovsk Krai shares its borders with Magadan Oblast in the north; with the Sakha Republic and Amur Oblast in the west; with the Jewish Autonomous Oblast, China (Heilongjiang), and Primorsky Krai in the south; and is limited by the Sea of Okhotsk in the east. In terms of area, it is the fourth-largest federal subject within Russia. Major islands include the Shantar Islands.
The Solve It! (with Newton's Second Law) Concept Builder provides learners plenty of practice using the F net = m•a equation to analyze situations involving unbalanced forces and accelerations. Much more than the usual Concept Builder, this activity demands that learners solve numerical problems. A word story problem with numerical ...
No daylight saving time (DST) in Khabarovsk. Sunrise and sunset time for Khabarovsk. Sunrise: 6:08 AM. Sunset: 7:53 PM. Day length: 13h 44m 50s. Solar noon: 1:01 PM. See the monthly sunrise, sunset, and twilight table for Khabarovsk. Time and time zones. The length of a solar day is determined by the time that it takes for the Earth to complete ...
The Newton's Second Law Concept Builder is shown in the iFrame below. There is a small hot spot in the top-left corner. Clicking/tapping the hot spot opens the Concept Builder in full-screen mode. Use the Escape key on a keyboard (or comparable method) to exit from full-screen mode. There is a second hot-spot in the lower-right corner of the ...