Answer:
A) 14.75
B) 3.36Kj
C) 2384.2k
D) 117.6kW
E) 57.69%
Explanation:
Attached is the full solutions.
Answer:
0.25 J/K
Explanation:
Given data in given question
heat (Q) = 100 J
temperature (T) = 400 K
to find out
the change in entropy of the given system
Solution
we use the entropy change equation here i.e
ΔS = ΔQ / T ...................a
Now we put the value of heat (Q) and Temperature (T) in equation a
ΔS is the entropy change, Q is heat and T is the temperature,
so that
ΔS = 100/400 J/K
ΔS = 0.25 J/K
Answer:
the pressure gradient in the x direction = -15.48Pa/m
Explanation:
- The concept of partial differentiation was used in the determination of the expression for u and v.
- each is partially differentiated with respect to x and the appropriate substitution was done to get the value of the pressure gradient as shown in the attached file.
Answer:
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Explanation:
there you go lad