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mash [69]
3 years ago
10

A student pulls a box across a horizontal floor at a constant speed of 4.0 meters per second by exerting a constant horizontal f

orce of 45 newtons. approximately how much work does the student do against friction in moving the box 5.5 meters across the floor?
Physics
2 answers:
ruslelena [56]3 years ago
8 0
Work = Force multiplied by the distance(or displacement)
Rainbow [258]3 years ago
5 0

Answer:

247.5N

Explanation:

A student pulls a box across a horizontal floor at a constant speed of 4.0 meters per second by exerting a constant horizontal force of 45 newtons. approximately how much work does the student do against friction in moving the box 5.5 meters across the floor?

DEFINITION OF TERMS

work done is the product of force and distance. if the student pulls the box without covering a distance, e asnt done any form of work.

force is tat which tends to change a body's state of rest or uniform motion  in a straight line.

w=f*d

45*5.5

247.5Nm

the frictional force between the object and the ground will be in the opposite direction to the force applied

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We're going to multiply the time it took for you to hear thunder (3.5 seconds) by the speed of sound in air (340 m/s)

3.5 x 340 = 1190

The lightning bolt was 1,190 meters away.
3 0
3 years ago
If the mass of an object is 10 Kg and it experiences a Net force of 60 Newtons. What is the acceleration?
Solnce55 [7]

Answer:

6 m/s^2

Explanation:

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3 0
2 years ago
What is the net force on these free body diagrams?
pentagon [3]

A

The horizontal force cancels out. The two 4Ns go in opposite directions. So they don't affect the outcome.

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Answer 4 N up

B

The horizontal and vertical forces cancel out. Each gives 3N - 3N =0

The net force is 0

C

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7 0
2 years ago
Read 2 more answers
If the speed at which an object moves through a fluid increases, will the size of the frictional force that acts on it increase,
Alla [95]

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increase

Explanation:

5 0
3 years ago
Four point charges are placed at the corners of a square. Each charge has the identical value +Q. The length of the diagonal of
kvv77 [185]

Answer:V_{net}=4\frac{kQ}{a}    

Explanation:

Given

charge on each Particle is Q

Length of diagonal of the square is 2a

therefore distance between center and each charge is \frac{2a}{2}=a

Electric Potential of charged Particle is given by

For First Charge

V_1=\frac{kQ}{a}

V_2=\frac{kQ}{a}

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V_4=\frac{kQ}{a}

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V_{net}=4\times \frac{kQ}{a}

V_{net}=4\frac{kQ}{a}            

7 0
2 years ago
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