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3 years ago

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Two point charges attract each other with an electric force of magnitude F. If one charge is reduced to one-third its original v

alue and the distance between the charges is doubled, what is the resulting magnitude of the electric force between them?a) 3F/2b) F/3c) F/6d) F/12e) 3F/4

1 answer:

Pavel [41]3 years ago

3 0

Answer:

New force, $F'=\dfrac{F}{12}$

Explanation:

Given that, two point charges attract each other with an electric force of magnitude F. It is given by :

$F=k\dfrac{q_1q_2}{r^2}$

If one charge is reduced to one-third its original value and the distance between the charges is doubled such that,

$q_1'=\dfrac{q_1}{3}$ , $r'=2r$

$F'=k\dfrac{q_1'q_2'}{r'^2}$

$F'=k\dfrac{(q_1/3)q_2}{(2r)^2}$

$F'=\dfrac{F}{12}$

So, the electric force between them is reduced to (1/12). Hence, the correct option is (d).

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A projectile is launched horizontally from a height of 8.0 m. The projectile travels 6.5 m before hitting the ground.

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You can find

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3) speed when it hits the ground

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y = yo - Vyo - gt^2 / 2

=> yo - y = gt^2 / 2

1) time to hit the ground

=> 8.0 = g t^2 / 2 => t^2 = 8.0m * 2 / 9.81 m/s^2 = 1.631 s^2

=> t = √1.631 s^2 = 1.28 s

2) initial velocity

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3) speed when it hits the ground

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3 years ago

Is a lawn mower a drag force OR friction force

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Answer:

Pretty sure it's friction force

Explanation:

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The phase speed vp for deep ocean waves is given by vp = g/ω where ω = 2π/T and T is the period of the wave; g is the accelerati

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Answer:

The group speed is $\dfrac{v_{p}}{2}$ .

Explanation:

Given that,

Phase speed $v_{p}=\dfrac{g}{\omega}$

$\omega=\dfrac{2\pi}{T}$

Where, T = period of the wave

g =acceleration due to gravity

We need to calculate the group speed

Using formula of phase speed

$v_{p}=\dfrac{\omega}{k}$

Put the value of $v_{p}$

$\dfrac{g}{\omega}=\dfrac{\omega}{k}$

$\omega=\sqrt{gk}$ ....(I)

We need to calculate the group speed

Using formula of group speed

$v_{g}=\dfrac{d\omega}{dk}$

$v_{g}=\dfrac{d}{dk}(\sqrt{gk})$

On differentiating

$v_{g}=\dfrac{1}{2}\sqrt{\dfrac{g}{k}}$

Put the value of k from equation (I)

$v_{g}=\dfrac{1}{2}\sqrt{\dfrac{g}{\dfrac{\omega^2}{g}}}$

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$v_{g}=\dfrac{v_{p}}{2}$

Hence, The group speed is $\dfrac{v_{p}}{2}$ .

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3 years ago

Consider two point charges located on the x axis: one charge, q1 = -11.5nC , is located at x1 = -1.685m ;the second charge, q2 =

Nata [24]

Answer:

$F_{t3}=-3.76\ 10^{-5}\ N$

The net force is pointed to the left side

Explanation:

Electrostatic Force

The formula to compute the electrostatic force between two point charges was introduced by Charles Coulomb and it's expressed as

$\displaystyle F=\frac{k\ q_1\ q_2}{r^2}$

Where q1 and q2 are the magnitudes of the charges, r is the distance between them and k is a proportionality constant

The data provided in the problem is

$\displaystyle x_1=-1,685\ m$

$\displaystyle q_1=-11.5\ nC$

$\displaystyle q_2=40\ nC$

$\displaystyle q_3=55\ nC$

$\displaystyle x_3=-1.195\ m$

The distance between the charge 1 and 3 is

$\displaystyle d_{13}=1.685-1.195$

$\displaystyle d_{13}=0.49\ m$

The distance between charges 2 and 3 is

$\displaystyle d_{23}=1.195\ m$

Now, let's compute the force exerted by q1 on q3

$\displaystyle F_{13}=\frac{k\ q_1\ q_3}{d_{13}^2}$

$\displaystyle F_{13}=\frac{9\ 10^9\ 11.5\ 10^{-9}\ 55\ 10^{-9}}{0.49^2}$

$\displaystyle F_{13}=2.37\ 10^{-5}\ N$

This value is the scalar magnitude of the force, we'll take care of the directions later. The force by q2 on q3 is

$\displaystyle F_{23}=\frac{k\ q_2\ q_3}{d_{23}^2}$

$\displaystyle F_{23}=\frac{9\ 10^{9}\ 40\ 10^{-9}\ 55\ 10^{-9}}{1.195^2}$

$\displaystyle F_{23}=1.39\ 10^{-5}\ N$

Given that q1 is negative and q3 is positive, q1 attracts q3 to the left, so F13 is in the negative direction. The charge q2 is positive and repels q3 to the left, thus F23 is also negative. This leads us to compute the total force on q3 as

$\displaystyle F_{t3}=-F_{13}-F_{23}$

$\boxed{F_{t3}=-3.76\ 10^{-5}\ N}$

The net force is pointed to the left side

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3 years ago