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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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ex 1) The sailboat moves north, while the wind moves from southeast. This means the angle between the direction of the boat and the wind is 45^{\circ}.

Calling F the force of the wind, and d=15~km=15000~m the distance covered by the boat, the work done by the wind is:
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The mechanical advantage (MA) of the ramp is the ratio between the output load (W) and the input force (F). The output load is the weight of the load, mg, therefore:
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Finally, the efficiency \epsilon of the ramp is the ratio between the output energy and the work done. The output energy is simply the potential energy (Ep) of the load, which is mgh, where h is the height of the ramp. The work done W is the product between the input force, F, and the displacement of the load, which is the length of the ramp: Fd. Therefore:
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ex 3) the graph is missing

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