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Nutka1998 [239]
3 years ago
15

You toss a conductive open ring of diameter d = 1.75 cm up in the air. The ring is flipping around a horizontal axis at a rate o

f 6.25 flips per second. One flip is a full rotation. At your location, the earth's magnetic field is B e = 30.5 μT. What is the maximum emf induced in the ring?
Physics
1 answer:
Mama L [17]3 years ago
8 0

Answer:

The maximum emf induced in the ring

= (2.882 × 10⁻⁷) V

Explanation:

According to the law of electromagnetic induction, the emf induced in the ring is given by

E = N BA w sin wt

The maximum emf induced is

E = N BA w

B = 30.5 μT = (30.5 × 10⁻⁶) T

A = (πD²/4)

D = 1.75 cm = 0.0175 m

A = (π×0.0175²/4) = 0.000240625 m²

Nw = 2π × 6.25 = 39.29 rad/s

E = 30.5 × 10⁻⁶ × 0.000240625 × 39.29

E = (2.882 × 10⁻⁷) V

Hope this Helps!!!

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mario62 [17]

Answer:

W = 290.7 dynes*cm

Explanation:

d = 1/5 cm = 0.2 cm

The force is in function of the depth x:

F(x) = 1000 * (1 + 2*x)^2

We can expand that as:

F(x) = 1000 * (1 + 4*x + 4x^2)

F(x) = 1000 + 4000*x + 4000*x^2

Work is defined as

W = F * d

Since we have non constant force we integrate

W = \int\limits^{0.2}_{0} {(1000 + 4000*x + 4000*x^2)} \, dx

W = [1000*x + 2000*x^2 + 1333*X^3] evaluated between 0 and 0.2

W = 1000*0.2 + 2000*0.2^2 + 1333*0.2^3 - 1000*0 - 2000*0^2 - 1333*0^3

W = 200 + 80 + 10.7 = 290.7 dynes*cm

3 0
3 years ago
The source of the earth’s magnetic field is found in the _____.
telo118 [61]
The source of the earth's magnetic field is found in the ;LIQUID MAGNETIZED IRON IN THE CORE OF THE EARTH. The earth's magnetic field is believed to be generated very deep down in the earth's core as a result of molten iron which it contains. 
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3 years ago
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A uniform electric field exists in the region between two oppositely charged plane parallel plates. A proton is released from re
valentinak56 [21]

Answer:

a) 17.33 V/m

b) 6308 m/s

Explanation:

We start by using equation of motion

s = ut + 1/2at², where

s = 1.2 cm = 0.012 m

u = 0 m/s

t = 3.8*10^-6 s, so that

0.012 = 0 * 3.8*10^-6 + 0.5 * a * (3.8*10^-6)²

0.012 = 0.5 * a * 1.444*10^-11

a = 0.012 / 7.22*10^-12

a = 1.66*10^9 m/s²

If we assume the electric field to be E, and we know that F =qE. Also, from Newton's law, we have F = ma. So that, ma = qE, and E = ma/q, where

E = electric field

m = mass of proton

a = acceleration

q = charge of proton

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E = 2.77*10^-18 / 1.6*10^-19

E = 17.33 V/m

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v = u + at

v = 0 + 1.66*10^9 * 3.8*10^-6

v = 6308 m/s

5 0
2 years ago
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Fittoniya [83]

Answer:

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Idk if this helps but plz mark brainliest if it does

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gayaneshka [121]

Answer:

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Explanation:

3 0
3 years ago
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