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Doss [256]
3 years ago
5

A 53-N force is needed to keep a 50.0-kg box sliding across a flat surface at a constant velocity. What is the coefficient of ki

netic friction between the box and the floor?0.11 a)0.10 b)0.13 c)0.09 d)0.11
Physics
2 answers:
earnstyle [38]3 years ago
5 0

The weight of the box is (mass) x (gravity) = (50 kg) x (9.8m/s²) = 490 newtons.

If the box is sliding at constant speed, and not speeding up or slowing down,
that means that the horizontal forces on it add up to zero. 

Since you're pushing on it with 53N in <em><u>that</u></em> direction, friction must be pulling
on it with 53N in the <u><em>other</em></u> direction.

 The 53N of friction is (the weight) x (the coefficient of kinetic friction).

                                                  53N  =  (490N) x (coefficient).

Divide each side by  490N :  Coefficient = (53N) / (490N)  =  0.1082 .

Rounded to the nearest hundredth, that's    <em>0.11 </em>.      (choice 'd')


Tems11 [23]3 years ago
4 0
The weight of the box is (mass) x (gravity) = (50 kg) x (9.8m/s²) = 490 newtons.
If the box is sliding at constant speed, and not speeding up or slowing down, <span>that means that the horizontal forces on it add up to zero.</span>
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hjlf

Answer:

I= 3.5 amps

Explanation:

Step one:

given data

rating of resistor R= 8 ohms

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Required

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Yet this power is also given by

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I= \sqrt{\frac{P}{R} }

substitute

I= \sqrt{\frac{100}{8} }\\\\I=\sqrt{12.5}\\\\I= 3.5 amps

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