Answer:
(d) 2 pF
Explanation: the charge on capacitor is given by the expression
Q=CV
where Q=charge
C=capacitance
V=voltage across the plate of the capacitor
here we have given Q=500 pF, V=250 volt
using this formula C=
=500××
=2×
=2 pF
What quantity measures the effect of change?
2 answers:
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Answer:
The exit temperature is 293.74 K.
Explanation:
Given that
At inlet condition(1)
P =80 KPa
V=150 m/s
T=10 C
Exit area is 5 times the inlet area
Now
If consider that density of air is not changing from inlet to exit then by using continuity equation
So
m/s
Now from first law for open system
Here Q=0 and w=0
When air is treating as ideal gas
Noe by putting the values
So the exit temperature is 293.74 K.