Answer:
pretty sure its B if it isnt im so so sorry
Explanation:
Yup displacement covered /time taken. bcz both velocity and displacement are vectors
Answer:
<em>2.15 mV</em>
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Explanation:
The complete question is
You have a coil of wire with 17 turns each of 1.5 cm radius. You place the plane of the coil perpendicular to a 0.50-TB? field produced by the poles of an electromagnet.
Find the magnitude of the average induced emf in the coil when the magnet is turned off and the field decreases to 0 T in 1.9 s .
Radius of the coil = 1.5 cm = 0.015 m
number of turns = 17
Initial magnetic field
= 0.50 T
Final magnetic field
= 0 T
time taken dt = 1.9 s
Area pf the coil = π
= 3.142 x
= 7.1 x 10^-4 m^2
magnetic flux = BA
initial flux =
A = 0.5 x 7.1 x 10^-4 = 3.55 x 10^-4 Wb
Final flux =
A = 0 x 7.1 x 10^-4 = 0 Wb
change in flux dФ = 3.55 x 10^-4 - 0 = 3.55 x 10^-4 Wb
Induced EMF E = NdФ/dt = (17 x 3.55 x 10^-4)/2.8 = 2.15 x 10^-3 V
==> <em>2.15 mV</em>
Answer:
ΔV = 20.1 V
Explanation:
As the positive plates are connected to each other, the capacitors are connected in parallel, so the total system load is the sum of the charges on each capacitor.
Q = Q₁ + Q₂
The charge on each capacitor is
Q₁ = C₁ ΔV₁
Q₁ = 24 10⁻⁶ 25
Q₁ = 6.00 10⁻⁴ C
Q₂ = C₂ ΔV₂
Q₂ = 13 10⁻⁶ 11
Q₂ = 1.43 10⁻⁴ C
The total set charge is
Q = (6 + 1.43) 10⁻⁴
Q = 7.43 10⁻⁴ C
The equivalent capacitance is
C_eq = C₁ + C₂
C_eq = (24 + 13) 10⁻⁶
C_eq = 37 10⁻⁶ F
Let's use the relationship to find the voltage
Q = C_eq ΔV
ΔV = Q / C_eq
ΔV = 7.43 10⁻⁴ / 37 10⁻⁶
ΔV = 2.008 10¹
ΔV = 20.1 V
This voltage is constant in the combination so it is also the voltage in capacitor C1