Answer:
a) before immersion
C = εA/d = (8.85e-12)(25e-4)/(1.31e-2) = 1.68e-12 F
q = CV = (1.68e-12)(255) = 4.28e-10 C
b) after immersion
q = 4.28e-10 C
Because the capacitor was disconnected before it was immersed, the charge remains the same.
c)*at 20° C
C = κεA/d = (80.4*)(8.85e-12)(25e-4)/(1.31e-2) = 5.62e-10 F
V = q/C = 4.28e-10 C/5.62e-10 C = 0.76 V
e)
U(i) = (1/2)CV^2 = (1/2)(1.68e-12)(255)^2 = 5.46e-8 J
U(f) = (1/2)(5.62e-10)(0.76)^2 = 1.62e-10 J
ΔU = 1.62e-10 J - 5.46e-8 J = -3.84e-8 J
Answer:
8.2N
Explanation:
I took the test and got it right :)
In physics, a force is said to do work<span> if, when acting, there is a displacement of the point of application in the direction of the force. It is expressed as </span><span>Work done = force (N) X distance (m). From the problem statement, the distance traveled is zero. Therefore, the work done is zero as well.</span>
Answer:
a) W=85.225 kW
b) 
Explanation:
First, consider the energy balance for the compressor: The energy that enters to the system (W and enthalpy of the feed flow) is equal to the energy that goes out from it (Heat Q and enthalpy of the exit flow):
Consider the enthalpy data from van Wylen 6th edition, Table B.2.2. According to that, 
, 
So, the power input to the compressor is:
b) The differential entropy change dS for a reversible heat transfer dQ at a temperature T is:
This equation can be integrated if the heat transfer surface temperature remains constant, which is the case, giving as a result:
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