Answer:
It is<em> impossible</em> to construct a machine which produces the <em>work output greater than the work input.</em>
Let us consider the II law of thermodynamics.
According to Kelvin Plank's statement any engine/machine does not give hundred percent efficiency. And violating the PMM-II(Perpetual motion of machine II kind), Always some amount of energy transferred to the sink or surroundings.
Therefore
W(ouput) = Q₁-Q₂
There are many reasons to lower the work output, just for an example friction between the mating parts reduces the work output.
Explanation:
F net of sled = Tension force by rope - Kinetic friction between ground.
F normal of sled = mg = (67kg)(9.81kg/m^2) = 657.27N.
Kinetic friction = 0.18 (I cannot see the value) * Normal force of sled = 0.18 * 657.27N = 118.31N
So F net of sled = 800N - 118.31N = 681.69N.
(I cannot see what the question is asking for, please check on your own!)
<span>Nothing happens to the pitch of a cell phone ring when the amplitude
of a sound wave increases.
Pitch and amplitude are both characteristics of a wave, but they're not
connected, and they don't influence each other.</span>
Answer:
a. 572Btu/s
b.0.1483Btu/s.R
Explanation:
a.Assume a steady state operation, KE and PE are both neglected and fluids properties are constant.
From table A-3E, the specific heat of water is
, and the steam properties as, A-4E:

Using the energy balance for the system:

Hence, the rate of heat transfer in the heat exchanger is 572Btu/s
b. Heat gained by the water is equal to the heat lost by the condensing steam.
-The rate of steam condensation is expressed as:

Entropy generation in the heat exchanger could be defined using the entropy balance on the system:

Hence,the rate of entropy generation in the heat exchanger. is 0.1483Btu/s.R