A fisherman sees his boat moving up and down, owing to waves on the water. It takes 2.5 s for the boat to travel from its highes
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Answer: A 100-lb child stands on a scale while riding in an elevator. Then, the scale reading approaches to 100lb, while the elevator slows to stop at the lowest floor
Explanation: To find the correct answer, we need to know more about the apparent weight of a body in a lift.
<h3>What is the apparent weight of a body in a lift?</h3>- Consider a body of mass m kept on a weighing machine in a lift.
- The readings on the machine is the force exerted by the body on the machine(action), which is equal to the force exerted by the machine on the body(reaction).
- The reaction we get as the weight recorded by the machine, and it is called the apparent weight.
- Here we have given with the actual weight of the body as 100lbs.
- This 100lb child is standing on the scale or the weighing machine, when it is riding .
- During this condition, the acceleration of the lift is towards downward, and thus, a force of ma .
- There is also<em> mg </em>downwards and a normal reaction in the upward direction.
- when we equate both the upward force and downward force, we get,
i.e. during riding the scale reads a weight less than that of actual weight.
- When the lift goes slow and stops the lowest floor, then the acceleration will be approaches to zero.
Thus, from the above explanation, it is clear that ,when the elevator moves to the lowest floor slowly and stops, then the apparent weight will become the actual weight.
Learn more about the apparent weight of the body in a lift here:
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Part a) The work done by the gas on the bullet is the integral of the force in dx, where x is the distance covered by the bullet inside the barrel with respect to the origin:
By substituting the length of the barrel, L=0.540 m, we find the total work done by the gas on the bullet:
part b) The resolution of the problem is the same, we just have to use the new length of the barrel (L=0.95 m) inside the final formula, and we find the new value of the work:
By substituting the length of the barrel, L=0.540 m, we find the total work done by the gas on the bullet:
part b) The resolution of the problem is the same, we just have to use the new length of the barrel (L=0.95 m) inside the final formula, and we find the new value of the work: