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.
<h3>How to solve the question?</h3>
- 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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Answer:
A. 4,9 m/s2
B. 2,0 m/s2
C. 120 N
Explanation:
In the image, 1 is going to represent the monkey and 2 is going to be the package. Let a_mín be the minimum acceleration that the monkey should have in the upward direction, so the package is barely lifted. Apply Newton’s second law of motion:
If the package is barely lifted, that means that T=m_2*g; then:
Solving the equation for a_mín, we have:
Once the monkey stops its climb and holds onto the rope, we set the equation of Newton’s second law as it follows:
For the monkey:
For the package:
The acceleration a is the same for both monkey and package, but have opposite directions, this means that when the monkey accelerates upwards, the package does it downwards and vice versa. Therefore, the acceleration a on the equation for the package is negative; however, if we invert the signs on the sum of forces, it has the same effect. To be clearer:
For the package:
We have two unknowns and two equations, so we can proceed. We can match both tensions and have:
Solving a, we have
We can then replace this value of a in one for the sums of force and find the tension T:
Answer:
ms⁻¹
Explanation:
= diameter of merry-go-round = 4 m
= radius of merry-go-round = = = 2 m
= moment of inertia = 500 kgm²
= angular velocity of merry-go-round before ryan jumps = 2.0 rad/s
= angular velocity of merry-go-round after ryan jumps = 0 rad/s
= velocity of ryan before jumping onto the merry-go-round
= mass of ryan = 70 kg
Using conservation of angular momentum
ms⁻¹