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mafiozo [28]
3 years ago
7

A 25.0 kg pickle is accelerated from rest through a distance of 6.0m in 4.0s across a level floor . If the friction force betwee

t the pickle and the floor is 3.8N, what is the work done to move the object
Physics
1 answer:
SIZIF [17.4K]3 years ago
8 0
Add the KE increase and the work done against friction.

The final velocity is twice the average, or 3.0 m/s
The final KE is (1/2)*25*3^2 = 112.5 J

The friction work done is 6*3.8 = 22.8 J 
 hope this is correct
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4 0
3 years ago
If he leaves the ramp with a speed of 35.0 m/s and has a speed of 33.0 m/s at the top of his trajectory, determine his maximum h
nadezda [96]

Answer:

H = 6.93 m

Explanation:

given data

velocity v = 35 m/s

horizontal component Vx = 33 m/s

solution

we get here maximum height so first we get vertical component here that is express as

Vy = \sqrt{v^2- Vx^2}        .........................1

put here value

Vy = \sqrt{35^2- 33^2}

Vy = 11.66 m/s

and

now we get height

H = \frac{Vy^2}{2g}        .............................2

put here value

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7 0
2 years ago
What does it mean when it asks how many Macromolecules are in a diet?
Alona [7]
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3 0
3 years ago
7. CALCULATE: How much work is done in each of
kifflom [539]
<h3>Answer</h3>

a. 8J

b. 90J

<h3>Notes formula</h3>

W = F × s

W = Work done (J)

F = Force (N)

s = distance (m)

<h3>Known that </h3>

a. F = 4N

s = 2m

b. F = 30N

s = 3m

<h3>Question</h3>

a. W = ..?

b. W = ..?

<h3>Way to do it</h3>

a. W = F × s

= 4N × 2m = 8J

b. W = F × s

= 30N × 3m = 90J

<em>#Moderators please don't be mean, dont delete my answers just to get approval from your senior or just to get the biggest moderation daily rank.</em>

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A ball is thrown horizontally with a velocity of 12 m/s. How
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Answer:

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2 years ago
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