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natta225 [31]
3 years ago
12

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.
Physics
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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Answer:

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You are missing the first part of the problem. This is an example, but it will give you the idea of how to solve yours with your data.

The first part is like this:

<em>A      4.0 cm  diameter parallel plate capacitor has a  0.44 m  m    gap. What is the displacement current in the capacitor if the potential difference across the capacitor is increasing at 500,000 V/s?</em>

Now with this, we can solve the problem.

In order to do this, we need to use the following expression:

q = CV (1)

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q: charge of plate or capacitor (In coulombs)

V: voltage in Volts.

However, we need is the current, and we have data of potential difference, so, all we have to do is divide the expression between time so:

q/t = CV/t

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C = 8.85x10^-12 * π * (0.04/2)² / 0.00046 = 2.42x10^-11 F

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