Answer:
I=9.6×e^{-8} A
Explanation:
The magnetic field inside the solenoid.
B=I*500*muy0/0.3=2.1×e ^-3×I.
so the total flux go through the square loop.
B×π×r^2=I×2.1×e^-3π×0.025^2
=4.11×e^-6×I
we have that
(flux)'= -U
so differentiating flux we get
so the inducted emf in the loop.
U=4.11×e^{-6}×dI/dt=4.11×e^-6×0.7=2.9×e^-6 (V)
so, I=2.9×e^{-6}÷30
I=9.6×e^{-8} A
Answer:
In refrigeration cycle heat transfer from inside refrigeration
In heat pump cycle heat transfer from environment
Explanation:
heat cycle is mechanical process use for cool the temperature but
In refrigeration heat transfer from inside of refrigeration that decrease temperature of refrigerator and in heat pump it decrease temperature negligible as compare to refrigerator
Answer:
b) False
Explanation:
Work done by a system is not a property because it doesn't define the system's state. Work is mechanical energy exchanged across the system's boundaries.
Answer:
(a) T₂ =747.5 and K= 474.5 °C (b) 330.178 kJ/kg
Explanation:
Solution
T₁ = 35°C = 308
the first step to take is to Use the Table A-17: Ideal gas properties for air:
Now,
At T₁ = 308 K
V₁ = 217.67 + [308-305/310-305] (221.25 -217.67)
So,
V₁ =219.818 kJ/kg
Thus,
Vr₁ = 596 + [308-305/310-305] (572.3 - 596)
= 581.78
so,
Vr₂/Vr₁ = 1/10
Vr₂ =58.178
Applying Table A-17, at Vr₂ = 58.178
Then,
(a) T₂ = 740 + [58.178 - 59.82/57.63 -59.82] (750 -740)
T₂ = 747.5 and K = 474.5 °C
V₂ =544.02 + [58.178 - 59.82/57.63 -59.82] (551.99 - 544.02)
so,
V₂ =549.996 kJ/kg
Hence,
(b) q - w = v₂ -v₁
= 0 -w = 549.996- 219.818
w = 330.178 kJ/kg
