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ella [17]
3 years ago
8

A uniform-density 7 kg disk of radius 0.21 m is mounted on a nearly frictionless axle. Initially it is not spinning. A string is

wrapped tightly around the disk, and you pull on the string with a constant force of 44 N through a distance of 0.9 m. Now what is the angular speed?
Physics
2 answers:
Thepotemich [5.8K]3 years ago
6 0

Answer:

22.65 rad/s

Explanation:

m = 7 kg, r = 0.21 m, F = 44 N, d = 0.9 m

Moment of inertia, I = 1/2 mr^2 = 0.5 x 7 x 0.21 x 0.21 = 0.15435 kg m^2

Work done, W = F x d = 44 x 0.9 = 39.6 J

according to the work energy theorem

Change in the kinetic energy of rotation = Work done

1/ 2 x I x ω^2 = W

0.5 x 0.15435 x ω^2 = 39.6

ω^2 = 513.12

ω = 22.65 rad/s

GREYUIT [131]3 years ago
5 0

Answer:

\omega = 22.67 rad/s

Explanation:

Here we can use energy conservation

As per energy conservation conditions we know that work done by external source is converted into kinetic energy of the disc

Now we have

W = \frac{1}{2}I\omega^2

now we know that work done is product of force and displacement

so here we have

W = F.d

W = (44 N)(0.9 m) = 39.6 J

now for moment of inertia of the disc we will have

I = \frac{1}{2}mR^2

I = \frac{1}{2}(7 kg)(0.21^2)

I = 0.154 kg m^2

now from above equation we will have

39.6 = \frac{1}{2}(0.154)\omega^2

\omega = 22.67 rad/s

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2. The hydrogen atom has quantized energy levels.

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What is oxygen an example of? a mixture, an element, a compound, or a solution
soldi70 [24.7K]

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7 0
3 years ago
A 1,250 W electric motor is connected to a 220 Vrms, 60 Hz source. The power factor is lagging by 0.65. To correct the pf to 0.9
Black_prince [1.1K]

Answer:

C = 46.891 \mu F

Explanation:

Given data:

v = 220 rms

power factor = 0.65

P = 1250 W

New power factor is 0.9 lag

we knwo that

s = \frac{P}{P.F} < COS^{-1} 0.65

S = \frac{1250}{0.65} < 49.45

s = 1923.09 < 49.65^o

s = [1250 + 1461 j] vA

P.F new = cos [tan^{-1} \frac{Q_{new}}{P}]

solving for Q_{new}

Q_{new} = P tan [cos^{-1} P.F new]

Q_{new} = 1250 [tan[cos^{-1}0.9]]

Q_{new} = 605.40 VARS

Q_C = Q - Q_{new}

Q_C = 1461 - 605.4 = 855.6 vars

Q_C = \frac[v_{rms}^2}{xc} =v_{rms}^2 \omega C

C = \frac{Q_C}{ v_{rms}^2 \omega}

C = \frac{855.6}{220^2 \times 2\pi \times 60}

C = 4..689 \times 10^{-5} Faraday

C = 46.891 \mu F

6 0
3 years ago
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