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:
5 I think will be none of the above and 6 could be all of the above
I need a picture plz I don’t know what to answer.
The chemical reaction causes electricity to flow through the terminals to the load attached. Some of the acid in the battery remains on the plates as it flows through. When the battery is recharged the acid is returned to the liquid solution to provide more power later.
When placing the piece of aluminium in water, the level of water will rise by an amount equal to the volume of the piece of aluminum.
Therefore, we need to find the volume of that piece.
Density can be calculated using the following rule:
Density = mass / volume
Therefore:
volume = mass / density
we are given that:
the density = 2.7 g / cm^3
the mass = 16 grams
Substitute in the equation to get the volume of the piece of aluminum as follows:
volume = 16 / 2.7 = 5.9259 cm^3
Since the water level will rise to an amount equal to the volume of aluminum, therefore, the water level will rise by 5.9259 cm^3
