When the body said to be in equilibrium
2 answers:
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Answer:
a.
b.
Explanation:
<u>Given:</u>
- Velocity of the particle, v(t) = 3 cos(mt) = 3 cos (0.5t) .
The acceleration of the particle at a time is defined as the rate of change of velocity of the particle at that time.
At time t = 3 seconds,
<u>Note</u>:<em> The arguments of the sine is calculated in unit of radian and not in degree.</em>
<h2>(b):</h2>
The velocity of the particle at some is defined as the rate of change of the position of the particle.
For the time interval of 2 seconds,
The term of the left is the displacement of the particle in time interval of 2 seconds, therefore,
It is the displacement of the particle in 2 seconds.
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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Answer:
The force has been reduced by 8018 N
Explanation:
The impulse exerted on the car during the crash is equal to the product of the force exerted and the duration of the collision, and it is also equal to the change in momentum of the car. So we can write:
where:
F is the force exerted on the car
is the duration of the collision
m = 1400 kg is the mass of the car
is the change in velocity of the car
We can re-write the equation as
In the 1st collision, the time is 1.5 seconds, so the force is
In the 2nd collision, the time is increased to 2.2 seconds, so the force is
Therefore, the force has been reduced by: