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3 years ago

Can someone help with these

Physics

1 answer:

yaroslaw [1]3 years ago

4 0

One

Givens

delta B = 0.20 T/s
A = 0.07 m^2
R = 3.5 ohms

Formula

Φ = ΔB*A

e = Φ

Solution (first part)

e = 0.2 * 0.07

e = 0.014 emf

Solution (second part)

i = e/R

i = 0.014 / 3.5

i = 4 * 10^-3

i = 4 ma

Answer

Two

Givens

N = 200 turns

Φ = 30 degrees

Delta B = 0.45 T/s

phi = 30 degrees

r = 0.06 meters

Formula

e = -N * delta B * A * Cos(phi)

Solution

e = -200 * 0.45 (pi r^2) * Cos(30)

e = - 200 * 0.45 * (3.14 * 0.06^2) * cos(30)

e = 0.881 emf

Answer

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The half-life of Co-55 is 175 hours. How much of a 4000 g of Cobalt-55 sample would be left after 525 hours?

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We know that whatever amount we start with, half of it decays and forms atoms of other elements in 175 hours. So in order to figure out how much is left after 525 hours, we'll need to know how many half-lifes pass in that amount of time.

Well, (525 divided by 175) is exactly 3 half-lifes. So this will be easy.

-- After 1 half-life . . .

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Another way:

After 1 half-life, 1/2 is left.

After 2 half-lifes, 1/4 is left.

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3 years ago

The current in a wire varies with time according to the relationship 1=55A−(0.65A/s2)t2. (a) How many coulombs of charge pass a

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Answer:

(A) Q = 321.1C (B) I = 42.8A

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(a)Given I = 55A−(0.65A/s2)t²

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To get an expression for Q we integrate with respect to t.

So Q = ∫I×dt =∫[55−(0.65)t²]dt

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3 years ago

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