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Mama L [17]
3 years ago
9

a 1 m length of string is wrapped around a solid disk, of mass .25 kg and a radius of .30m, mounted on a frictionless axle. the

string is pulled with a tension force of 20 N. what is the angular velocity of the disk after the string is pulled off
Physics
1 answer:
Nezavi [6.7K]3 years ago
8 0

Answer:

60 rad/s

Explanation:

∑τ = Iα

Fr = Iα

For a solid disc, I = ½ mr².

Fr = ½ mr² α

α = 2F / (mr)

α = 2 (20 N) / (0.25 kg × 0.30 m)

α = 533.33 rad/s²

The arc length is 1 m, so the angle is:

s = rθ

1 m = 0.30 m θ

θ = 3.33 rad

Use constant acceleration equation to find ω.

ω² = ω₀² + 2αΔθ

ω² = (0 rad/s)² + 2 (533.33 rad/s²) (3.33 rad)

ω = 59.6 rad/s

Rounding to one significant figure, the angular velocity is 60 rad/s.

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Calculate the intensity of current flowing through a computer that consumes 180W and operates at 120 V.
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C) 1.5 A

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A 25 N force stretches a spring 280 cm. What was the spring constant? ​
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Answer:

  1. F= 25N
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How do you do this question?
umka2103 [35]

Explanation:

The moment of inertia of each disk is:

Idisk = 1/2 MR²

Using parallel axis theorem, the moment of inertia of each rod is:

Irod = 1/2 mr² + m (R − r)²

The total moment of inertia is:

I = 2Idisk + 5Irod

I = 2 (1/2 MR²) + 5 [1/2 mr² + m (R − r)²]

I = MR² + 5/2 mr² + 5m (R − r)²

Plugging in values:

I = (125 g) (5 cm)² + 5/2 (250 g) (1 cm)² + 5 (250 g) (5 cm − 1 cm)²

I = 23,750 g cm²

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A runner traveling with an initial velocity of 1.1 m/s accelerates at a constant rate of 0.8 m/s2fora time of 2.0 s.(a).What is
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Answer:

The final velocity of the runner at the end of the given time is 2.7 m/s.

Explanation:

Given;

initial velocity of the runner, u = 1.1 m/s

constant acceleration, a = 0.8 m/s²

time of motion, t = 2.0 s

The velocity of the runner at the end of the given time is calculate as;

v = u + at

where;

v is the final velocity of the runner at the end of the given time;

v = 1.1 + (0.8)(2)

v = 2.7 m/s

Therefore, the final velocity of the runner at the end of the given time is 2.7 m/s.

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