Answer:
<u>Option-(A)</u>
Explanation:
<u>Typical applications for the high carbon steels includes the following;</u>
It is heat treatable, relatively large combinations of mechanical characteristics. Typical applications: railway wheels and tracks, gears, crankshafts, and machine parts.
Answer:
sorry i dont understand the answer
Explanation:
but i think its a xd jk psml lol
Answer:
a) Zero
b) the rate of entropy generation in the system's universe = ds/dt = 0.2603 KW/K
Explanation:
a) In steady state
Net rate of Heat transfer = net rate of heat gain - net rate of heat lost
Hence, the rate of heat transfer = 0
b) In steady state, entropy generated
ds/dt = - [ Qgain/Th1 + Qgain/Th2 - Qlost/300 K]
Substituting the given values, we get –
ds/dt = -[5/1500 + 3/1000 – (5+3)/300]
ds/dt = - [0.0033 + 0.003 -0.2666]
ds/dt = 0.2603 KW/K
I believe the answer is: A. Passive heating and cooling.
Answer:
V = 56.8 mV
Explanation:
When a current I flows across a circuit element, if we assume that the dimensions of the circuit are much less than the wavelength of the power source creating this current, there exists a fixed relationship between the power dissipated in the circuit element, the current I and the voltage V across it, as follows:
P = V*I
By definition, power is the rate of change of energy, and current, the rate of change of the charge Q, so we can replace P and I, as follows:
E/t = V*q/t ⇒ E = V*Q
Solving for V:
V = E/Q = 94.2 mJ /1.66 C = 56.8 mV
