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Nezavi [6.7K]
3 years ago
14

A Circular loop in the plane of a paper lies in a 0.65 TT magnetic field pointing into the paper. The loop's diameter changes fr

om 17.5 cmcm to 6.6 cmcm in 0.48 ss .
a) Determine the direction of the induced current and justify your answer.b) Determine the magnitude of the average induced emf.c) If the coil resistance is 2.5 Ω, what is the average induced current?
Physics
1 answer:
Kobotan [32]3 years ago
6 0

Answer:

Explanation:

magnetic filed, B = 0.65 T

initial diameter, d = 17.5 cm

final diameter, d' = 6.6 cm

time, t = 0.48 s

(a) According to Lenz's law, the direction of induced current is clockwise.

(b) Let e is the induced emf.

initial area, A = π r² = 3.14 x 0.0875 x 0.0875 = 0.024 m²

final area, A' = π r'² = 3.14 x 0.033 x 0.033 = 0.00342 m²

change in area, ΔA = A - A' = 0.024 - 0.00342 = 0.02058 m²

The magnitude of induced emf is given by

e=\frac{d\phi }{dt}

e=B\frac{\Delta A }{dt}

e = 0.65 x 0.02058 / 0.48

e = 0.028 V

(c) R = 2.5 ohm

i = e / R

i = 0.028 / 2.5

i = 0.011 A

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3 years ago
If 270 watts of power is used in 42 seconds, how much work was done<br>​
navik [9.2K]

Answer: W = 11340J

Explanation:

Hey there! I will give the following steps, if you have any questions feel free to ask me in the comments below.

So this is the Formula: Power = Work / Time.

<u>Step 1:</u><em><u> Find the Formula</u></em>

P = W / T

<em><u> </u></em>

<u>Step 2: </u><u><em>Make W the subject of the equation.</em></u>

W = PT

<u>Step 3:</u><u> </u><u><em>Given.</em></u>

P = 270 Watts, T = 42 seconds

<u>Step 4:</u><u><em> Substitute these values into equation 2 .</em></u>

W = 270(42)

<u>Step 5:</u><u> </u><u><em>Simplify.</em></u>

W = 11340J

The amount of work done was 11340.

~I hope I helped you! :)~

4 0
4 years ago
A 3.00-kg object has a velocity 1 6.00 i ^ 2 2.00 j ^2 m/s. (a) what is its kinetic energy at this moment? (b) what is the net w
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(b) The new velocity is 8.00 m/s on the x-axis and 4.00 m/s on the y-axis, so the magnitude of the new velocity is
v= \sqrt{(8.00 m/s)^2+(4.00 m/s)^2}=8.94 m/s
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K= \frac{1}{2}mv^2= \frac{1}{2}(3.00 kg)(8.94 m/s)^2=120 J

So, the work done on the object is the variation of kinetic energy of the object:
W=\Delta K=120 J-60 J=60 J
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