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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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Answer:

Explanation:

Given

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3 years ago

In 1.71 minutes, a ski lift raises four skiers at constant speed to a height of 148 m. The average mass of each skier is 62.9 kg

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Answer:

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3 years ago

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a) The work done by the tree is $-1.44\cdot 10^5 J$

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Explanation:

a)

According to the work-energy theorem, the work done on the car is equal to the change in kinetic energy of the car. Therefore, we can write:

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W is the work done on the car

m is the mass of the car

u is its initial speed

v is its final speed

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m = 2000 kg

u = 12.0 m/s

v = 0 (since the car comes to a stop, after the crash)

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b)

The work done by the tree on the car can also be rewritten as

$W=Fd$

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F is the force applied on the car

d is the displacement of the car during the collision

In this situation, we have:

$W=-1.44\cdot 10^5 J$ is the work done

$d=50.0 cm = 0.50 m$ is the displacement of the car during the collision

Solving the equation for F, we find the force exerted by the tree on the car:

$F=\frac{W}{d}=\frac{-1.44\cdot 10^5 J}{0.50}=-2880 N$

Where the negative sign means the force is applied opposite to the direction of motion of the car. Therefore, the magnitude of the force applied is 2880 N.

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