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Crank
3 years ago
9

A horizontal force of 150 N is used to push a 40.0-kg packing crate a distance of 6.00 m on a rough horizontal surface. If the c

rate moves at constant speed, find (a) the work done by the 150-N force and (b) the coefficient of kinetic friction between the crate and surface.
Physics
1 answer:
icang [17]3 years ago
3 0

Answer:

a. 900 J

b. 0.383

Explanation:

According to the question, the given data is as follows

Horizontal force = 150 N

Packing crate = 40.0 kg

Distance = 6.00 m

Based on the above information

a. The work done by the 150-N force is

W = F x = \mu N x = \mu\ m\ g\ x

W = 150 \times 6

= 900 J

b. Now the coefficient of kinetic friction between the crate and surface is

\mu = \frac {F}{m\timesg}

= \frac{150}{40\times 9.8}

= .383

We simply applied the above formulas so that each one part could calculate

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ss7ja [257]

PART a)

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Answer:

The  time elapsed at the spacecraft’s frame is less that the time elapsed at earth's  frame

Explanation:

From the question we are told that

The distance between earth and Retah is  d = 20 \ light \ hours =  20 * 3600 *  c =  72000c \ m

Here c is the peed of light with value c =  3.0*10^8 m/s

The time taken to reach Retah from earth is  t =  25 \ hours  =  25 * 3600 =90000 \ sec

The velocity of the spacecraft is mathematically evaluated  as

     v_s =  \frac{d }{t}

substituting values

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The time elapsed in the spacecraft’s frame is mathematically evaluated as

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substituting value

       T  =  90000 *  \sqrt{ 1 -  \frac{[2.4*10^{8}]^2}{[3.0*10^{8}]^2} }

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Answer:

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6 0
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