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SIZIF [17.4K]
3 years ago
7

A heat engine operating at steady state delivers a power output of 15.5 hp. The engine receives energy by heat transfer in the a

mount of Qh = 30.5 kJ during each cycle of operation from a high temperature thermal reservoir at Th = 940 K. The system is executing 50 cycles per minute. Determine the amount of work that is delivered during each cycle in kilojoules
Physics
1 answer:
Anna007 [38]3 years ago
7 0

Answer:

w_cycle = 13.68 KJ / cycle

Explanation:

Given:

- Steady power out-put W_out = 15.5 hp

- System executes 50 cycles/min

Find:

Determine the amount of work that is delivered during each cycle in kilo-joules

Solution:

- First we will convert the steady output into KW as follows:

                                 W_out = 15.5 hp

The conversion factor is 1.36 KW per hp.

                                 W_out = 15.5 [hp] * [KW] / 1.36 [hp]

                                 W_out = 11.4 [KW] or [KJ/s]

- Given that there are 50 cycles in 60 sec. We will use direct proportionality to calculate the number of cycles per second:

                       50 cycles   -----------> 60 s

                          x cycles   -----------> 1 s

                                 x = 50/60 = 0.8333 cycles /s

- Next we will compute the amount of work done per cycle:

                                 w_cycle = W_out / x

                                 w_cycle = 11.4 [KJ/s] * [s] / 0.8333 [cycles]

                                 w_cycle = 13.68 KJ / cycle

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In this case, the horizontal velocity of the rocket starts from the acceleration, so if its velocity drops to zero,

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A 10n force is applied to a 25kg mass to slide it across a frictional surface. What is the acceleration of the mass?
klasskru [66]

Answer: a = 0.4m/s^2 - 9.8*c where c is the coefficient of kinetic friction of the surface

Explanation: We know that, by the second Newton's law, a = F/m

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Then, if the surface is frictionless, the total force applied in the object is 10N, and the mass of the object is 25kg, so the acceleration is:

a =10N/25kg = 0.4m/s^2.

But if the surface is frictional, there will be a force of friction applied in the mass (this depends on the coefficient of friction and the weight of the mass), this means that the acceleration will be reduced.

If = -(9.8*25)*c

where c is a number that is bigger than 0 and smaller than 1, is called the coefficient of kinetic friction.

So the total force is now:

F = (10 - 9.8*25*c)

Then, the acceleration in a frictional surface is equal to:

a = (10 - 9.8*25*c)/25 = 0.4m/s^2 - 9.8*c

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3 years ago
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Which of the following is a conductor?<br><br> copper<br> water<br> aluminum<br> all of the above
hjlf

Answer:

D all of the above

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3 years ago
Suzy drops a rock from the roof of her house. Mary sees the rock pass her 2.9 m tall window in 0.134 sec. From how high above th
timama [110]

Answer:

Explanation:

Given

length of window h=2.9\ m

time Frame for which rock can be seen is \Delta t=0.134\ s

Suppose h is height above which rock is dropped

Time taken to cover h+2.9 is t_1

so using equation of motion

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and from equation t_1-t_2=0.134\ s

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t_1+t_2=4.416\ s

and t_1=t_2+\Delta t

so t_2+\Delta t+t_2=4.416

2t_2+0.134=4.416

t_2=0.5\times 4.282

t_2=2.141\ s

substitute the value of t_2 in equation 2

h=0.5\times 9.8\times (2.141)^2

h=22.46\ m

                                                     

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