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saveliy_v [14]
3 years ago
8

A potter's wheel is rotating around a vertical axis through its center at a frequency of 2.0 rev/s . The wheel can be considered

a uniform disk of mass 5.0 kg and diameter 0.36 m . The potter then throws a 3.1-kg chunk of clay, approximately shaped as a flat disk of radius 8.0 cm , onto the center of the rotating wheel. What is the frequency of the wheel after the clay sticks to it? Ignore friction.
Physics
1 answer:
weeeeeb [17]3 years ago
8 0

Answer:

ωf =1.78 rev/s

Explanation:

it is a non elastic collision, so conservation of angular momentum is used.  and there id no friction

I₁ωi₁ + I₂ωi₂ = (I₁ + I₂)ωf

ωi₁ = 2.0 rev/s*(2π) rad/rev = 4.0π rad/s

assume ωi₂ = 0

Sol the equation to wf

ωf =  (½m₁r₁²)ωi₁ /  (½m₁r₁² + ½m₂r₂²)

ωf = (m₁r₁²)*ωi₁  /  (m₁r₁² + m₂r₂²)

ωf = (5.0kg*(0.18m)²)4.0 π rad/s / ((5.0kg*(0.18m)²) + (3.1*(0.08m)²))

ωf =11.19 rad/s/ 2π

ωf =1.78 rev/s

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Accelerating charges radiate electromagnetic waves. Calculate the wavelength of radiation produced by a proton of mass mp moving
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3 0
3 years ago
A uniformly charged solid disk of radius R = 0.45 m carries a uniform charge density of σ = 175 μC/m². A point P is located a di
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Explanation:

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R = 0.45 m

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  • The Electric Field Strength E of a uniformly solid disk of charge at distance a perpendicular to disk is given by:

                                  E = 2*pi*k*o * (1 - \frac{a}{\sqrt{a^2 + R^2} })\\

part a)

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E = 2*pi*8.99*10^9*175*10^-6 * (1 - \frac{0.75}{\sqrt{0.75^2 + 0.45^2} })\\\\E = 9885021.285*(0.1425070743)\\\\E = 1408.685 KN/C

part b)

Since, R >> a, we can approximate a / R = 0 ,

Hence, E simplified relation becomes:

E = 2*pi*k*o * (1 - \frac{a/R}{\sqrt{a^2/R^2 + 1} })\\\\E = 2*pi*k*o * (1 - \frac{0}{\sqrt{0 + 1} })\\\\E = 2*pi*k*o

E = σ / 2*e_o

part c)

Since, a >> R, we can approximate. that the uniform disc of charge becomes a single point charge:

Electric Field strength due to point charge is:

E = k*δ*pi*R^2 / a^2  

Since, R << a, Surface area = δ*pi

Hence,

E = (k*δ*pi/a^2)

 

6 0
3 years ago
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