Explanation:
After some time t the current does not passing through the circuit
=>so the back emf is zero
=>here the inductor opposes decay of the circuit
- Ldi/dt = Ri
di/dt = - R/Li
di/i = - R/Ldt
now we applying the integration on both sides
log i=-R/Lt+C
here t=0=>i=io
Log io=C
=>Log i=-R/L*t + Log io
logi-Log io=-R/L*t
Log[i/io]=-R/L*t
i/io=e^-Rt/L
i=ioe^-Rt/L
the option D is correct
Answer:
The equation of equilibrium at the top of the vertical circle is:
\Sigma F = - N - m\cdot g = - m \cdot \frac{v^{2}}{R}
The speed experimented by the car is:
\frac{N}{m}+g=\frac{v^{2}}{R}
v = \sqrt{R\cdot (\frac{N}{m}+g) }
v = \sqrt{(5\,m)\cdot (\frac{6\,N}{0.8\,kg} +9.807\,\frac{kg}{m^{2}} )}
v\approx 9.302\,\frac{m}{s}
The equation of equilibrium at the bottom of the vertical circle is:
\Sigma F = N - m\cdot g = m \cdot \frac{v^{2}}{R}
The normal force on the car when it is at the bottom of the track is:
N=m\cdot (\frac{v^{2}}{R}+g )
N = (0.8\,kg)\cdot \left(\frac{(9.302\,\frac{m}{s} )^{2}}{5\,m}+ 9.807\,\frac{m}{s^{2}} \right)
N=21.690\,N
G
has the SI units
m
3
k
g
⋅
s
2
I assume that the ball is stationary (v=0) at point B, so its total energy is just potential energy, and it is equal to 7.35 J.
At point A, all this energy has converted into kinetic energy, which is:
And since K=7.35 J, we can find the velocity, v:
