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cupoosta [38]
3 years ago
11

A person is standing on a scale placed on the floor of an elevator. At time t1, the elevator is at rest and the reading on the s

cale is 500N. At a later time t2, the person is still standing on the scale and the reading on the scale is 400N . Based on the scale readings, which of the following statements about the motion of the elevator could be true at time t2?
a. The elevator is moving upward at constant speed.
b. The elevator is moving upward and slowing down.
c. The elevator is moving downward at constant speed.
d. The elevator is moving downward and slowing down.
e. The elevator is at rest.
Physics
2 answers:
sergiy2304 [10]3 years ago
7 0

the following statements about the motion of the elevator could be true at time t2 is

c. The elevator is moving downward at constant speed.

Explanation:

  • Under given situation, the elevator is at rest and the reading on the scale is 500N.
  • After some time t2, the person is still standing on the scale and the reading on the scale is 400N .
  • It is because if you stand on a scale in an elevator which is accelerating upward, your body will feel heavier because the elevator's floor pressure which presses harder on your feet, and this is why the scale will show a higher reading than the time when the elevator is at rest.
  • Similarly on the other hand, when the elevator accelerates downward, your body will feel lighter. The force which is exerted by the scale is called as the apparent weight; which means it does not change with constant speed.
  • Applying Newton's second law, which concludes about this particular statement about force exerted on a body when at rest and when in motion.

Gennadij [26K]3 years ago
7 0

Answer:

id say a

Explanation:

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Answer:

VR = \frac{Radius of wheel}{Radius of axle}

Explanation:

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Answer:

141.78 ft

Explanation:

When speed, u = 44mi/h, minimum stopping distance, s = 44 ft = 0.00833 mi.

Calculating the acceleration using one of Newton's equations of motion:

v^2 = u^2 + 2as\\\\v = 0 mi/h\\\\u = 44 mi/h\\\\s = 0.00833 mi\\\\=> 0^2 = 44^2 + 2 * a * 0.00833\\\\=> 1936 = -0.01666a\\\\a = -116206.48 mi/h^2 or -14.43 m/s^2

Note: The negative sign denotes deceleration.

When speed, v = 79mi/h, the acceleration is equal to when it is 44mi/h i.e. -116206.48 mi/h^2

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v^2 = u^2 + 2as\\\\v = 0 mi/h\\\\u = 79 mi/h\\\\a = -116206.48 mi/h\\\\=> 0^2 = 79^2 + (2 * -116206.48 * s)\\\\6241 = 232412.96s\\\\s = \frac{6241}{232412.96} \\\\s = 0.0268531 mi = 141.78 ft

The minimum stopping distance is 141.78 ft.

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