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Varvara68 [4.7K]
3 years ago
5

You are designing an elevator for a hospital. The force exerted on a passenger by the floor of the elevator is not to exceed 1.4

6 times the passenger's weight. The elevator accelerates upward with constant acceleration for a distance of 2.2 m and then starts to slow down.What is the maximum speed of the elevator?
Physics
1 answer:
vovikov84 [41]3 years ago
4 0

Answer:

Final velocity of the elevator will be 4.453 m/sec

Explanation:

Let mass is m

Acceleration due to gravity is g m/sec^2

Distance s = 2.2 m

As the elevator is moving upward so net force on elevator

F=mg+ma

So according to question

1.46mg=mg+ma

0.46 mg = ma

a = 0.46 g

a = 0.46×9.8 = 4.508 m/sec^2

Initial velocity of elevator is 0 m/sec

From third equation of motion

v_f^2=v_i^2+2as

v_f^2=0^2+2\times 4.508\times 2.2

v_f=4.453m/sec

So final velocity of the elevator will be 4.453 m/sec

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Calculate (a) the torque, (b) the energy, and (c) the average power required to accelerate Earth in 4.0 days from rest to its pr
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<h2>Answer:</h2>

Torque = <em>2.05 x 10²⁸ Nm</em>

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<em />

t = 4 days = 4 x 24 x 3600 seconds = 345600 seconds

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=> α = (7.29 × 10⁻¹⁰) / (3.456)

=> α = 2.11 × 10⁻¹⁰ rad/s²

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τ = 9.71 x 10³⁷ x 2.11 × 10⁻¹⁰

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E = \frac{1}{2} x 9.71 x 10³⁷ x 7.29 × 10⁻⁵

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(iii) The average power P, is given by;

P = E / t

P = 3.54 x 10³³ / 345600

P = 1.02 x 10²⁸ W

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