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posledela
4 years ago
12

A NASA spacecraft measures the rate of at which atmospheric pressure on Mars decreases with altitude. The result at a certain al

titude is: Convert to .
Physics
1 answer:
Vlada [557]4 years ago
5 0

Answer:

= 4.21\times 10^{-10}J/cm^4

Explanation:

According to the given situation, the computation is shown below:-

1kPa=10^3Pa

So,

1Km = 10^5cm

1kPa=0.01Pa/Cm

1kPa=10^{-8}J/cm^4

Now as per the question it is given that

r = 0.0421KPa/Km

= 0.0421KPa/Km \times \frac{10^{-8}J/cm^4}{1KPa/Km}

after solving the above equation we will get

= 0.0421\times 10^{-8}J/cm^4

or we can write the above equation in this manner, that is

= 4.21\times 10^{-10}J/cm^4

Therefore we have solved the equation to reach the right answer.

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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:

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

     

7 0
4 years ago
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