Answer:
The amount of work done required to stretch spring by additional 4 cm is 64 J.
Explanation:
The energy used for stretching spring is given by the relation :
.......(1)
Here k is spring constant and x is the displacement of spring from its equilibrium position.
For stretch spring by 2.0 cm or 0.02 m, we need 8.0 J of energy. Hence, substitute the suitable values in equation (1).

k = 4 x 10⁴ N/m
Energy needed to stretch a spring by 6.0 cm can be determine by the equation (1).
Substitute 0.06 m for x and 4 x 10⁴ N/m for k in equation (1).

E = 72 J
But we already have 8.0 J. So, the extra energy needed to stretch spring by additional 4 cm is :
E = ( 72 - 8 ) J = 64 J
Applicable linear expansion equation:
ΔL = αΔTL
In which
ΔL = change in length, α = Linear expansion coefficient of steel, ΔT = change in temperature, L = original length
Therefore,
ΔL = 12*10^-6*(18.5-(-3))*1410 = 0.36378 m
Answer:
The inductance of the inductor is 35.8 mH
Explanation:
Given that,
Voltage = 120-V
Frequency = 1000 Hz
Capacitor 
Current = 0.680 A
We need to calculate the inductance of the inductor
Using formula of current


Put the value of Z into the formula

Put the value into the formula


Hence, The inductance of the inductor is 35.8 mH
C and E Gravitational and Chemical energy