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timama [110]
2 years ago
5

A 60 kilogram astronaut weighs 96 newtons on the surface of the moon. calculate the acceleration due to gravity on the moon.

Physics
2 answers:
Charra [1.4K]2 years ago
8 0

Answer:

Acceleration due to gravity on moon, 1.6\ m/s^2

Explanation:

Mass of the astronaut, m = 60 kg

Weight of astronaut on the surafce of moon, W = 96 N

The weight of astronaut is given by :

W = m g

g=\dfrac{W}{m}

g=\dfrac{96\ N}{60\ kg}

g=1.6\ m/s^2

So, the acceleration due to gravity on the surface of moon is 1.6\ m/s^2

Also, the acceleration due to gravity on the surface of moon is (1/6)th of the acceleration due to gravity on the surface of earth.

Hence, this is the required solution.

34kurt2 years ago
6 0
Given: Mass m = 60 Kg

           Weight  W = 96 N

Required: Acceleration due to gravity, g = ?

Formula:  W = mg

                g = W/g

                g = 96 Kg.m/s²/60 Kg   (note: this is the derive unit for Newton "N")

                g = 1.6 m/s²

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Answer:1.27

Explanation:

Given

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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
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Answer:

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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

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

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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

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substituting values

       W = 63000 J

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The Carnot efficiency is mathematically represented as

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

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

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        \eta _a  =   \frac{W}{E_h}

substituting values

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

         \eta _a  = 0.46

     

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