The emf induced in the second coil is given by:
V = -M(di/dt)
V = emf, M = mutual indutance, di/dt = change of current in the first coil over time
The current in the first coil is given by:
i = i₀
i₀ = 5.0A, a = 2.0×10³s⁻¹
i = 5.0e^(-2.0×10³t)
Calculate di/dt by differentiating i with respect to t.
di/dt = -1.0×10⁴e^(-2.0×10³t)
Calculate a general formula for V. Givens:
M = 32×10⁻³H, di/dt = -1.0×10⁴e^(-2.0×10³t)
Plug in and solve for V:
V = -32×10⁻³(-1.0×10⁴e^(-2.0×10³t))
V = 320e^(-2.0×10³t)
We want to find the induced emf right after the current starts to decay. Plug in t = 0s:
V = 320e^(-2.0×10³(0))
V = 320e^0
V = 320 volts
We want to find the induced emf at t = 1.0×10⁻³s:
V = 320e^(-2.0×10³(1.0×10⁻³))
V = 43 volts
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Answer:
100 Watts
Explanation:
<u>These equations are needed to work out the answer:</u>
- power= work done/ time taken
- work done= force* distance
- force= mass* acceleration
force: 10 kg* 2m/s= 20
work done: 20* 10m=200
power: 200/2=100
Answer:

Explanation:
For this problem, we need to apply the formulas of constant accelerated motion.
To obtain the boat displacement we need to calculate the displacement because of the river flow and the displacement done because of the boat motor.
for the river:

for the boat:

So the final displacement is given by:
