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GREYUIT [131]
2 years ago
12

Can you solve the issue

Physics
1 answer:
loris [4]2 years ago
3 0

Answer:

The answer is  271.3 joules .

Explanation:

          Work = (weight) x (distance)

 Work = (50 lb) x (1 kg / 2.20462 lb) x (9.81 newton/kg)

                          x (4 feet) x (1 meter / 3.28084 feet)

          = (50 x 9.81 x 4) / (2.20462 x 3.28084)  newton-meter

          =        271.3 joules .

We don't need to know how long the lift took, unless we

want to know how much power he was able to deliver.

                  Power = (work) / (time)    

                              = (271.3 joule) / (5 sec)  =  54.3 watts .

________________________________________

The easy way:

        Work = (weight) x (distance)

                  = (50 pounds) x (4 feet)  =  200 foot-pounds

Look up (online) how many joules there are in 1 foot-pound.

There are  1.356 joules in 1 foot-pound.

So  200 foot-pounds = (200 x 1.356) = 271.2 joules.

credits original answer AL2006

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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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