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4 years ago

9

A 2 kg soccer ball is traveling 28.62m/s when it hits the wall and bounces off of the wall with a velocity of 20 m/s. If the wal

l exerted an average force of 652.36N on the ball, how long was the ball in contact with the wall?

1 answer:

Lesechka [4]4 years ago

5 0

Answer:

0.149 s or 0.15 s

Explanation:

let initially ball is moving towards left hence initial velocity = - 28.62 m/s

final velocity as ball moves right = +20 m/s

force = rate of change in momentum

force = mass × change in velocity / time

or time = mass × change in velocity / force

time = 2× ( 20 -( -28.62)) / 652.36

time = 2× ( 20 +28.62)) / 652.36

time = 2× 48 .62/652.36

time = 0.149 s or 0.15 s

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3 years ago

How much work was done by a hot air balloon to lift up a 100 Newton to a height of 300 meters?

Monica [59]

So, the work was done by that hot air-balloon is <u>30,000 J or 30 kJ</u>.

<h3>Introduction</h3>

Hi ! In this question, I will help you. <u>Work is the amount of force exerted to cause an object to move a certain distance from its starting point</u>. In physics, the amount of work will be proportional to the increase in force and increase in displacement. Amount of work can be calculated by this equation :

$\boxed{\sf{\bold{W = F \times s}}}$

With the following condition :

W = work (J)
F = force (N)
s = shift or displacement (m)

Now, the s (displacement) can be written as ∆h (altitude change) because the object move to vertical line. The formula can also be changed to:

$\boxed{\sf{\bold{W = F \times \Delta h}}}$

With the following condition :

W = work (J)
F = force (N)
$\sf{\Delta h}$ = change of altitude (m)

If an object has mass, then the object will also be affected by gravity. Always remember that F = m × g. So that :

$\sf{W = F \times \Delta h}$

$\boxed{\sf{\bold{W = m \times g \times \Delta h}}}$

With the following condition :

W = work (J)
m = mass of the object (kg)
g = acceleration of the gravity (m/s²)
$\sf{\Delta h}$ = change of altitude (m)

<h3>Problem Solving</h3>

We know that :

F = force = 100 N
$\sf{\Delta h}$ = change of altitude 300 m

What was asked :

W = work = ... J

Step by step :

$\sf{W = F \times \Delta h}$

$\sf{W = 100 \times 300}$

$\boxed{\sf{W = 30,000 \: J = 30 \: kJ}}$

<h3>Conclusion</h3>

So, the work was done by that hot air-balloon is 30,000 J or 30 kJ.

<h3>See More :</h3>

Work that he had done to lift object brainly.com/question/26341717
Converting work to potential energy brainly.com/question/26487284

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