A 0.245 kg ball is thrown straight up from 2.07 m above the ground. Its initial vertical speed is 8.00 m/s. A short time later,
1 answer:
Answer:
The work done by gravity is
Explanation:
The data given in the question is :
Mass is
Height from ground is
As we know , the work done is state function , it depends on initial and final position not on the path followed.
So, work done by gravity = change in potential energy
Work done = Initial potential energy - final potential energy
Insert values from question
Work done =
Work done =
So, work done =
Hence the work done by gravity is
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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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Answer: 52 kg skydiver: 9.09 m/s and 522.55 s
95 kg skydiver: 12.3 m/s and 386.2 s
Explanation: <u>Drag</u> <u>Force</u> is an opposite force when an object is moving in a fluid.
For skydivers, when falling through the air, the forces acting on it are gravitational and drag forces. At a certain point, drag force equals gravitational force, which is constant on any part of the planet, producing a net force that is zero. Since there is no net force, there is no acceleration and, consequently, velocity is constant. When that happens, the person reached the <u>Terminal</u> <u>Velocity</u>.
Drag Force and Velocity are proportional to the squared speed. So, terminal velocity is given by:
where
m is mass in kg
g is acceleration due to gravitational force in m/s²
ρ is density of the fluid in kg/m³
C is drag coefficient
A is area of the object in the fluid in m²
Calculating:
The 52kg skydiver has terminal velocity of:
9.09
The 95kg skydiver's terminal velocity is
12.3
The 52 kg and 95kg skydivers' terminal velocity are 9.09m/s and 12.3m/s, respectively.
The time each one will reach the floor will be:
52 kg at 9.09 m/s:
t = 522.5
95 kg at 12.3 m/s:
t = 386.2
The 52 kg and 95kg skydivers' time to reach the floor are 522.5 s and 386.2 s, respectively.