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Greeley [361]
3 years ago
11

A 13.4-mH inductor carries a current i =

{max}" align="absmiddle" class="latex-formula"> sin ωt, with I_{max} = 4.80 A and f = ω/2π = 60.0Hz. What is the self-induced emf as a function of time? (Express your answer in terms of t where e m f is in volts and t is in seconds. Do not include units in your expression.)

Physics
1 answer:
Digiron [165]3 years ago
6 0

The voltage across an inductor ' L ' is

V = L · dI/dt .

I(t) = I(max) sin(ωt)

dI/dt = I(max) ω cos(ωt)

V = L · ω · I(max) cos(ωt)

L = 1.34 x 10⁻² H

ω = 2π · 60 = 377 /sec

I(max) = 4.80 A

V = L · ω · I(max) cos(ωt)

V = (1.34 x 10⁻² H) · (377 / sec) · (4.8 A) · cos(377 t)

<em>V = 24.25 cos(377 t)</em>

V is an AC voltage with peak value of 24.25 volts and frequency = 60 Hz.

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change in momentum

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The doctor writes a prescription for na heparin 20,000 units in 500 ml n.s. infuse over 8 hours. what is the flow rate in ml/hr?
Ne4ueva [31]
I think this type of equation could be conducted in simple division equation since it does not involve drop rate.

we know that there is 500 ml of substance and should be infused within 8 hours period.

So the flow rate in ml/hr would be: 

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8 0
2 years ago
Distance is 80 seconds is 8 what is the average speed?
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8 0
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Compressing the air by squeezing the bottle was accompanied by a(n) ________ in the temperature of air inside the bottle
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3 0
3 years ago
A) 1.2-kg ball is hanging from the end of a rope. The rope hangs at an angle 20° from the vertical when a 19 m/s horizontal wind
Marat540 [252]

Answer:

Part a)

F_v = 4.28 N

Part B)

L = 1.02 m

Part C)

v = 1.25 m/s

Explanation:

Part A)

As we know that ball is hanging from the top and its angle with the vertical is 20 degree

so we will have

Tcos\theta = mg

T sin\theta = F_v

\frac{F_v}{mg} = tan\theta

F_v = mg tan\theta

F_v = 1.2\times 9.81 (tan20)

F_v = 4.28 N

Part B)

Here we can use energy theorem to find the distance that it will move

-\mu mg cos\theta L + mg sin\theta L = -\frac{1}{2}mv^2

(-(0.37)m(9.81) cos15 + m(9.81) sin15)L = - \frac{1}{2}m(1.4)^2

(-3.5 + 2.54)L = - 0.98

L = 1.02 m

Part C)

At terminal speed condition we know that

F_v = mg

bv^2 = mg

2.5 v^2 = 3.9

v = 1.25 m/s

7 0
2 years ago
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