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3 years ago

- A person is on an elevator that moves

Physics

1 answer:

Mademuasel [1]3 years ago

5 0

Answer:

The net force acting on the person is, F = 560 N

Explanation:

Given data,

The downward acceleration of the elevator, a = 1.8 m/s²

The weight of the person is, W = mg

= 686 N

Therefore the mass of the person,

m = W /g

= 686 / 9.8

= 70 kg

The net force acting on the person,

F = mg - ma

= 686 - 70 x 1.8

= 560 N

Hence, the net force acting on the person is, F = 560 N

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2) 2418 rad/s

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4) 36.3 m/s

Explanation:

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It can be calculated using the following suvat equation for angular motion:

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$\theta$ is the angular displacement

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In this problem we have:

t = 1.3 ms = 0.0013 s is the time elapsed

$\omega_i = 0$ is the initial angular velocity

$\alpha = 1.86\cdot 10^6 rad/s$ is the angular acceleration

Therefore, the final angular speed is:

$\omega_f = 0 + (1.86\cdot 10^6)(0.0013)=2418 rad/s$

The tangential acceleration is related to the angular acceleration by the following formula:

$a_t = \alpha r$

where

$a_t$ is the tangential acceleration

$\alpha$ is the angular acceleration

r is the distance of the point from the centre of rotation

Here we want to find the tangential acceleration of the tip of the claw, so:

$\alpha = 1.86\cdot 10^6 rad/s$ is the angular acceleration

r = 1.5 cm = 0.015 m is the distance of the tip of the claw from the axis of rotation

Substituting,

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$v=u+at$

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a is the tangential acceleration

t is the time elapsed

Here we have:

$a=27900 m/s^2$ (tangential acceleration)

u = 0 m/s (it starts from rest)

t = 1.3 ms = 0.0013 s is the time elapsed

Substituting,

$v=0+(27900)(0.0013)=36.3 m/s$

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