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jok3333 [9.3K]
3 years ago
12

A ring of radius 8 cm that lies in the ya plane carries positive charge of 2 μC uniformly distributed over its length. A particl

e of mass m that carries a charge of-2 oscillates about the center of the ring with an angular frequency of 22 rad/s Find the angular frequency of oscillation of the mass if the radius of the ring is doubled
Physics
1 answer:
zepelin [54]3 years ago
4 0

Answer:

The angular frequency of oscillation of the mass is 11 rad/s.

Explanation:

Given that,

Charge = 2μC

Radius R₁= 8 cm

Radius R₂ = 16 cm

Angular frequency = 22 rad/s

We need to calculate the angular frequency of oscillation of the mass

The electric field produced along x axis

E=\dfrac{kqx}{\sqrt{R^2+x^2}}

E=\dfrac{kqx}{\sqrt{R^2(1+\dfrac{x^2}{R^2})^2}}

E=\dfrac{kqx}{R^3}

The force on the mass is

F=Eq

F=\dfrac{kQqx}{R^3}....(I)

For,x<<R

Now, using centripetal force

F = \dfrac{mv^2}{r}

Put the value of F in equation (I)

\dfrac{mv^2}{r}=\dfrac{kQq}{R^3}

We know that,

v=r\omega

m\omega^2r=\dfrac{kQq}{R^3}

\omega^2=\dfrac{kQq}{mrR^3}

For, r<<R

\omega^2=\dfrac{kQq}{mR^3}

\omega=\sqrt{\dfrac{kQq}{mR^3}}

Here,

\omega\propto\sqrt{\dfrac{q}{R^3}}

The ratio of angular frequency

\dfrac{\omega}{\omega_{1}}=\sqrt{\dfrac{\dfrac{q}{R^3}}{\dfrac{q_{1}}{R_{1}^3}}}

\dfrac{\omega}{\omega'}=\sqrt{\dfrac{R^3\times2q}{2R^3\times q}}

\omega=\sqrt{\dfrac{8^3\times2\times2\times10^{-6}}{8\times8^3\times2\times10^{-6}}}\times\omega'

\omega=0.5\omega'

Put the value of

\omega=\dfrac{1}{2}\times22

\omega=11\ rad/s

Hence, The angular frequency of oscillation of the mass is 11 rad/s.

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Round the following off to required number of significant figures.
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A bubble, located 0.200 m beneath the surface in a glass of beer, rises to the top. The air pressure at the top is 1.01 ???? 10
SIZIF [17.4K]

Answer:

the ratio of the bubble’s volume at the top to its volume at the bottom is 1.019

Explanation:

given information

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\frac{V_{2} }{V_{1}} = \frac{P_{1} }{P_{2}}

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8 0
3 years ago
A train 471 m long is moving on a straight track with a speed of 75.1 km/h. The engineer applies the brakes at a crossing, and l
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Answer:

t = 37.6 s

Explanation:

As we know that train is initially moving with the speed

v_i = 75.1 km/h

now we know that

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v_f = 4.17 m/s

now we can use kinematics here

v_f^2 - v_i^2 = 2 a d

4.17^2 - 20.86^2 = 2 a(471)

a = -0.44 m/s^2

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4 0
2 years ago
A ray of light incident in air strikes a rectangular glass block of refractive index 1.5 at angle of incidence of 45.
Gwar [14]

Refraction is said to occur when there is a change in the speed of light.

<h3>What is the angle of refraction?</h3>

We know that refraction is said to occur when there is a change in the speed of light as it travels form one medium to another.

Given that the refractive index of the rectangular glass block is  1.5. The angle of refraction can be obtained by the use of the Snell's law;

n = sin i /sinr

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sini = angle of incidence

sin r = angle of refraction

sinr = sini/n

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b) Now;

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For the glass

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sin r = 0.392

r = 23 degrees

Learn more about angle of refraction:brainly.com/question/2660868

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4 0
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