Answer:
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Answer:
ΔU= 1.3922 KJ
Explanation:
Given that
Q= 2 KJ
Note- 1 .Heat added to the system taken as positive and heat leaving from the system is taken negative.
2. Work done on the system taken as negative and work done by the system taken as positive.
P =2 atm
V₁= 2 L
V₂=5 L
Work done by the gas W
W= P ( V₂- V₁)
W= 2 ( 5 - 2)
W= 6 atm.L
1 L·atm = 0.1013 kJ
W= 0.6078 KJ
From first law of thermodynamics
Q= W + ΔU
ΔU=Change in internal energy
2 = 0.6078 + ΔU
ΔU= 1.3922 KJ
Answer:
Part a)
Part b)
U = 111 J
Explanation:
As we know that the capacitor is of capacitance
now the maximum charge on it
Part a)
Oscillation frequency of the charge is given as
Part b)
Now the equation of charge oscillation is given as
now current in the circuit is given as
now at t = 1.35 ms we have
So the energy stored in inductor is given as
If you just type "<span>What is the chemical formula for mercury(I) nitrate?" into google you get the answer but HG(NO3)2 is the correct one.
sorry no one helped you in time hope you passed anyway</span>