A _ increases it decreases voltage in a power line
1 answer:
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The question is incomplete. Here is the complete question.
The image below was taken with a camera that can shoot anywhere between one and two frames per second. A continuous series of photos was combined for this image, so the cars you see are in fact the same car, but photographed at differene times.
Let's assume that the camera was able to deliver 1.3 frames per second for this photo, and that the car has a length of approximately 5.3 meters. Using this information and the photo itself, approximately how fast did the car drive?
Answer: v = 6.5 m/s
Explanation: The question asks for velocity of the car. Velocity is given by:
The camera took 7 pictures of the car and knowing its length is 5.3, the car's displacement was:
Δx = 7(5.3)
Δx = 37.1 m
The camera delivers 1.3 frames per second and it was taken 7 photos, so time the car drove was:
1.3 frames = 1 s
7 frames = Δt
Δt = 5.4 s
Then, the car was driving:
v = 6.87 m/s
The car drove at, approximately, a velocity of 6.87 m/s
Answer:
a) <em>2.278 x 10^-5 volts</em>
b) <em>1.139 x 10^-6 Ampere</em>
c) <em>2.59 x 10^-11 W</em>
Explanation:
The radius of the wire r = 2 mm = 0.002 m
the number of turns N = 200 turns
direction of the magnetic field ∅ = 25°
magnetic field strength B = 0.02 T
varying time = 2 sec
The cross sectional area of the wire =
==> A = 3.142 x = 1.257 x 10^-5 m^2
Field flux Φ = BA cos ∅ = 0.02 x 1.257 x 10^-5 x cos 25°
==> Φ = 2.278 x 10^-7 Wb
The induced EMF is given as
E = NdΦ/dt
where dΦ/dt = (2.278 x 10^-7)/2 = 1.139 x 10^-7
E = 200 x 1.139 x 10^-7 = <em>2.278 x 10^-5 volts</em>
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b) If the two ends are connected to a resistor of 20 Ω, the current through the resistor is given as
= E/R
where R is the resistor
= (2.278 x 10^-5)/20 = <em>1.139 x 10^-6 Ampere</em>
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<em> </em>c) power delivered to the resistor is given as
P = E
P = (1.139 x 10^-6) x (2.278 x 10^-5) = <em>2.59 x 10^-11 W</em>