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jasenka [17]
3 years ago
10

A car tries to tow a stranded van out of a ditch, and the car applies 300 N and pulls the van for 3 meters in less than a minute

before getting it safely back onto the road. How much work was done by the car on the car? The van on the car?
Physics
1 answer:
sergejj [24]3 years ago
8 0

Answer:

Work done, W = 900 Joules

Explanation:

It is given that,

Force applies by the car on the van, F = 300 N

The van pulls a distance of, d = 3 m

We need to find the work was done by the car on the car. We know that the product of force and distance is equal to the work done by the object i.e.

W=F\times d

W=300\ N\times 3\ m

W = 900 Joules

So, the work done by the car on the van is 900 Joules. Hence, this is the required solution.

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A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 79.6 m/s at ground level.
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Answer:

The rocket is motion above the ground for 44.7 s.

Explanation:

The equations for the height of the rocket are as follows:

y = y0 + v0 · t + 1/2 · a · t²

and, after the engine fails:

y = y0 + v0 · t + 1/2 · g · t²

Where:

y = height of the rocket at time t

y0 =  initial height

v0 = initial speed

t=  time

a = acceleration due to the engines

g = acceleration due to gravity

First, let´s calculate the time the rocket is being accelerated by the engines:

(The center of the framer of reference is located at the ground, y0 = 0).

y = y0 + v0 · t + 1/2 · a · t²

1190 m = 0 m + 79.6 m/s · t + 1/2 · 4.10 m/s² · t²

0 = 2.05  m/s² · t² + 79.6 m/s · t - 1190 m

Solving the quadratic equation:

t = 11.5 s  (the other value of t is discarded because it is negative).

At that time, the engines fail and the rocket starts to fall. The equation of the height will be:

y = y0 + v0 · t + 1/2 · g · t²

The initial velocity (v0) will be the velocity acquired during 11.5 s of acceleration plus the initial velocity of launch:

v = v0 + a · t

v = 79.6 m/s + 4.10 m/s² · 11.5 s

v = 126.8 m/s

Now, we can calculate the time it takes the rocket to reach the ground (y = 0) from a height of 1190 m:

y = y0 + v0 · t + 1/2 · g · t²

0 m = 1190 m + 126.8 m/s · t - 1/2 · 9.80 m/s² · t²

Solving the quadratic equation:

t = 33.2 s

Then, the total time the rocket is in motion is (33.2 s + 11.5 s) 44.7 s

 

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3 years ago
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Answer:

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The formula of weight:

W = Mass x Gravitational force

W = m x g

Given data:

Mass =1 kg

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W = 1kg x 9.8 ms-2 = 9.8 kgms-2 ( 1 kgms-2 = N)

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Define second class lever​
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Please find detailed explanation of second class levers below

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