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jeka57 [31]
2 years ago
6

A 63.3 kg wood board is resting on very smooth ice in the middle of a frozen lake. A 35.7 kg boy stands at one end of the board.

He walks from one end of the board to the other end with a velocity of 1.47 m/s relative to the ice in the positive direction. What is the velocity of the board relative to the ice?
Physics
1 answer:
Vlada [557]2 years ago
8 0

The velocity of the board relative to the ice is zero, since both are at rest.

<h3>What is relative velocity?</h3>

Relative velocity is the velocity of an object in relation to another reference object or point.

When two objects are travelling or moving with the same velocity in the same direction, the relative velocity one relative to the other is zero.

Also, when two objects are at rest, the relative velocity one relative to the other is zero.

Therefore, the velocity of the board relative to the ice is zero, since both are at rest.

Learn more about relative velocity at: brainly.com/question/24337516

#SPJ1

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An engine draws energy from a hot reservoir with a temperature of 1250 K and exhausts energy into a cold reservoir with a temper
dimulka [17.4K]

Answer:

The power output of this engine is  P =  17.5 W

The  the maximum (Carnot) efficiency is  \eta_c  = 0.7424

The  actual efficiency of this engine is  \eta _a  = 0.46

Explanation:

From the question we are told that

    The temperature of the hot reservoir is  T_h = 1250 \ K

      The temperature of the cold reservoir  is  T_c  =  322 \ K

     The energy absorbed from the hot reservoir is E_h  = 1.37 *10^{5} \ J

       The energy exhausts into  cold reservoir is  E_c  = 7.4 *10^{4} J

The power output is mathematically represented as

      P  =  \frac{W}{t}

Where t is the time taken which we will assume to be 1 hour =  3600 s  

W is the workdone which is mathematically represented as

      W =  E_h  -E_c

substituting values

       W = 63000 J

So

    P =  \frac{63000}{3600}

    P =  17.5 W

The Carnot efficiency is mathematically represented as

          \eta_c  =  1 - \frac{T_c}{T_h}

         \eta_c  =  1 - \frac{322}{1250}

         \eta_c  = 0.7424

The actual efficiency is mathematically represented as

        \eta _a  =   \frac{W}{E_h}

substituting values

         \eta _a  =  \frac{63000}{1.37*10^{5}}

         \eta _a  = 0.46

     

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