Answer:
C
Explanation:
If you asses the risks you can prepare for the worst
Given Information:
Inductance = L = 5 mH = 0.005 H
Time = t = 2 seconds
Required Information:
Current at t = 2 seconds = i(t) = ?
Energy at t = 2 seconds = W = ?
Answer:
Current at t = 2 seconds = i(t) = 735.75 A
Energy at t = 2 seconds = W = 1353.32 J
Explanation:
The voltage across an inductor is given as
The current flowing through the inductor is given by
Where L is the inductance and i(0) is the initial current in the inductor which we will assume to be zero since it is not given.
So the current at t = 2 seconds is
The energy stored in the inductor at t = 2 seconds is
Answer:
Enthalpy at outlet=284.44 KJ
Explanation:
We need to Find enthalpy of outlet.
Lets take the outlet mass m and outlet enthalpy h.
So from mass conservation
m=1+1.5+2 Kg/s
m=4.5 Kg/s
Now from energy conservation
By putting the values
So h=284.44 KJ
Answer:
The elastic modulus of the steel is 139062.5 N/in^2
Explanation:
Elastic modulus = stress ÷ strain
Load = 89,000 N
Area of square cross section of the steel bar = (0.8 in)^2 = 0.64 in^2
Stress = load/area = 89,000/0.64 = 139.0625 N/in^2
Length of steel bar = 4 in
Extension = 4×10^-3 in
Strain = extension/length = 4×10^-3/4 = 1×10^-3
Elastic modulus = 139.0625 N/in^2 ÷ 1×10^-3 = 139062.5 N/in^2