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3 years ago

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An elevator accelerates uniformly from rest to a speed of 2.5 m/s in 12s. What is the distance the elevator travels during this

time? must show work.

1 answer:

Aleks04 [339]3 years ago

7 0

We use the kinematic equation,

$s= u t +\frac{1}{2}a t^2$

Here, s is the distance traveled by elevator u is the initial velocity and t is time.

As elevator accelerates uniformly from rest to a speed of 2.5 m/s in 12 s, so

$a = \frac{v_{f}- v_{i} }{t} = \frac{2.5 m/s-0}{12 s} = 0.208 \ m/s^2$

Thus, the distance the elevator travels during 12 s time is

$s= 0+\frac{1}{2} 0.208 \ m/s^2(12)^2 = 14.97 m \simeq 15 m$

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3 years ago

(a) What is the fluid speed in a fire hose with a 9.00-cm diameter carrying 80.0 L of water per second? (b) What is the flow rat

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3 years ago

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LuckyWell [14K]

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