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3 years ago

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Which of the following changes would double the force between two charged particles?

1 answer:

soldier1979 [14.2K]3 years ago

4 0

Answer:

Doubling the amount of charge on one of the particles.

Explanation:

The force between two charges is given by :

$F=\dfrac{kq_1q_2}{r^2}$

Where

r is the distance between charges

or

$F\propto \dfrac{1}{r^2}$

On doubling the charge on one of the particle,

F' = 2F

So, the force gets doubled. Hence, the correct option is (d).

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Answer:

0.07756 m

Explanation:

Given mass of object =0.20 kg

spring constant = 120 n/m

maximum speed = 1.9 m/sec

We have to find the amplitude of the motion

We know that maximum speed of the object when it is in harmonic motion is given by $v_{max}=A\omega$ where A is amplitude and $\omega$ is angular velocity

Angular velocity is given by $\omega=\sqrt{\frac{k}{m}}$ where k is spring constant and m is mass

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WINSTONCH [101]

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3 years ago

A charged capacitor is connected to an ideal inductor to form an LC circuit with a frequency of oscillation f = 1.6 Hz. At time

IrinaK [193]

Answer:

$8.0\mu C$

Explanation:

We are given that

$f=1.6 Hz$

$q=3.0\mu C=3.0\times 10^{-6} C$

$1\mu C=10^{-6} C$

Current,I= $75\mu A=75\times 10^{-6} A$

$1\mu A=10^{-6} A$

We have to find the maximum charge of the capacitor.

Charge on the capacitor, $q=q_0cos\omega t$

$\omega=2\pi f=2\pi\times 1.6=3.2\pi rad/s$

$3\times 10^{-6}=q_0cos3.2\pi t$ ....(1)

$I=\frac{dq}{dt}=-q_0\omega sin\omega t$

$75\times 10^{-6}=-q_0(3.2\pi)sin3.2\pi t$ ....(2)

Equation (2) divided by equation (1)

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$3.2\pi t=tan^{-1}(-2.488)=-1.188rad$

$q_0=\frac{q}{cos\omega t}=\frac{3\times 10^{-6}}{cos(-1.188)}=8.0\times 10^{-6}=8\mu C$

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