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
We have,
- Jane mass is 55 kg
- His body covered with 700 nails all of them having a surface area of 1.00 mm² each = 700 × 1 = 700 mm² = 700/1000000 = 7/10000
We know that,
Let's calculate force as we already have area;
- F = ma
- F = 55 × 9.8 { Acceleration due to gravity }
- F = 539 N
Now, if should she would be on 700 nails then pressure will be;
- P = F/A
- P = 539/7 × 10000
- P = 5390000/7
- P = 770,000 Pascal
And if should would be on a 1 nail only,
- P = F/A
- P = 539/1 × 1000000
- P = 539000000 Pascal
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Answer:
V= 6.974 m/s
Explanation:
Component( box) weight acting parallel and down roof 88(sin39.0°)=55.4 N
Force of kinetic friction acting parallel and up roof = 18.0 N
Fnet force acting on tool box acting parallel and down roof
Fnet= 55.4 - 18.0
Fnet=37.4 N
acceleration of tool box down roof
a = 37.4(9.81)/88.0
a= 4.169 m/s²
d = 4.90 m
t = √2d/a
t= √2(4.90)/4.169
t= 1.662 s
V = at
V= 4.169(1.662)
V= 6.974 m/s
Answer:
Explanation:
Initial momentum is 1.5e6(3) = 4.5e6 kg•m/s
An impulse results in a change of momentum
The tug applied impulse is 12000(10) = 120000 N•s or 0.12e6 kg•m/s
The remaining momentum is 4.5e6 - 0.12e6 = 4.38e6 kg•m/s
The barge velocity is now 4.38e6 / 1.5e6 = 2.92 m/s
The tug applies 0.012e6 N•s of impulse each second.
The initial barge momentum will be zero in
t = 4.5e6 / 0.012e6 = 375 s or 6 minutes and 15 seconds
To stop the barge in one minute(60 s), the tug would have to apply
4.5e6 / 60 = 75000 N•s /s or 75 000 N