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Butoxors [25]
2 years ago
5

The experimenter fom the video rotates on his stool, this time holding his empty hands in his lap. You stand on a desk above him

and drop a long, heavy been bag straight down so that it lands across his lap, in his hands. what happens?
Physics
1 answer:
gtnhenbr [62]2 years ago
7 0

Answer:

Explanation:

The experimenter is rotating on his stool with angular velocity ω ( suppose )

His moment of inertia is I say

We are applying no torque from outside . therefore , the angular momentum will remain the same

Thus angular momentum L = I ω = constant

Thus we can say I₁ ω₁ = I₂ω₂ = constant

here I₁ is the initial moment of inertia and ω₁ is the initial angular velocity

Similarly I₂ is the final moment of inertia and ω₂ is the final angular velocity

When a been bag is dropped on his lap , his moment of inertia increases due to increase in mass

In the above equation, when moment of inertia increases , the angular velocity decreases . So its motion of rotation will decrease .

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A series RLC circuit with a resistance of 121.0 Ω has a resonance angular frequency of 5.1 ✕ 105 rad/s. At resonance, the voltag
Vadim26 [7]

Answer:

Explanation:

At resonance ω₀L = 1 / ω₀C , L is inductance and C is capacitance .

C = 1 /  ω₀²L , ω₀ = 5.1 x 10⁵ . ( given )

voltage over resistance = R I , R is resistance and I is current

voltage over inductance   =  Iω₀L

R I /  Iω₀L = 60 / 40

R / ω₀L = 3 / 2

L = 2 R / 3 ω₀

= 2 x 121 / 3 x 5.1 x 10⁵

= 15.81 x 10⁻⁵

C = 1 /  ω₀²L

= 1 / (5.1 x 10⁵)² x 15.81 x 10⁻⁵

= .002432 x 10⁻⁵

= 24.32 x 10⁻⁹ F

Let the angular frequency required be ω

Tan 45 = (ωL - 1 / ωC) / R

ωL - 1 / ωC = R

ω²LC - 1 = R ωC

ω²LC = 1 + R ωC

ω² x 15.81 x 10⁻⁵ x 24.32 x 10⁻⁹ = 1 + 121 x ω x 24.32 x 10⁻⁹

ω² x 384.5 x 10⁻¹⁴ = 1 + 2942.72 x10⁻⁹ω

ω² - 7.65 x 10⁶ ω - 1 = 0

ω = 7.65 x 10⁶

frequency = 7.65 x 10⁶ / 2π

= 1.22 x 10⁶ Hz

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3 years ago
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Romashka [77]
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iris [78.8K]
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fenix001 [56]

Answer:

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