Both are at the side of the spectrum that has the lower frequency
If the boat is i travling at 10 m/s and the river is 8.0 m/s the boats speed is 18.0 m/s
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
Answer:
d. 50 C
Explanation:
In this problem, we have to add 800 ml of water at 20 Celsius to 800 ml of water at 80 Celsius.
According to the 2nd law of thermodynamics, heat transfers from hot to cold temperature.
The quantity of both the different waters is equal so this makes it very easy. All we have to do is find the mean of both the temperatures:
Final temperature = (20 C + 80 C)/2
= 50 Celsius