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3 years ago

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Two electric charges, held a distance, dd, apart experience an electric force of magnitude, FF, between them. If one of the char

ges is doubled in magnitude while maintaining the same separation between the charges, what is the new magnitude of the force between them?

1 answer:

lorasvet [3.4K]3 years ago

3 0

Answer:

$F'=2F$

Explanation:

The Coulomb's law states that the magnitude of the electrostatic force between two charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them:

$F=\frac{kq_1q_2}{d^2}$

In this case, we have $q_1'=2q_1$ :

$F'=\frac{kq'_1q_2}{d^2}\\F'=\frac{k(2q_1)q_2}{d^2}\\F'=2\frac{kq_1q_2}{d^2}\\F'=2F$

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Two high-current transmission lines carry currents of 29.0 A and 78.0 A in the same direction and are suspended parallel to each

jarptica [38.1K]

Answer with Explanation:

We are given that

$I_1=29 A$

$I_2=78 A$

$d=38 cm=\frac{38}{100}=0.38 m$

$1 m=100 cm$

a.Length of segment,l=20 m

Magnetic force ,F= $\frac{2\mu_0I_1I_2 l}{4\pi d}$

$\frac{\mu_0}{4\pi}=10^{-7}$

Substitute the values

$F=\frac{10^{-7}\times 29\times 78\times 20}{0.38}=0.0119 N$

Hence, the magnetic force exert by each segment on the other=0.0119 N

b.We know that when current carrying in the wires are in same direction then the force will attract to each other.

Hence, the force will be attractive.

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3 years ago

Read 2 more answers

A 2 kg rubber ball is thrown at a wall horizontally at 3 m/s, and bounces back the way it came at an equal speed. A 2 kg clay ba

Lyrx [107]

Answer:

THE RUBBER BALL

Explanation:

From the question we are told that

The mass of the rubber ball is $m_r = 2 \ kg$

The initial speed of the rubber ball is $u = 3 \ m/s$

The final speed at which it bounces bank $v - 3 \ m/s$

The mass of the clay ball is $m_c = 2 \ kg$

The initial speed of the clay ball is $u = 3 \ m/s$

The final speed of the clay ball is $v = 0 \ m/s$

Generally Impulse is mathematically represented as

$I = \Delta p$

where $\Delta p$ is the change in the linear momentum so

$I = m(v-u)$

For the rubber is

$I_r = 2(-3 -3)$

$I_r = -12\ kg \cdot m/s$

=> $|I_r| = 12\ kg \cdot m/s$

For the clay ball

$I_c = 2(0-3)$

$I_c = -6 \ kg\cdot \ m/s$

=> $| I_c| = 6 \ kg\cdot \ m/s$

So from the above calculation the ball with the a higher magnitude of impulse is the rubber ball

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4 years ago

Which of the following are functions of both cells and living organisms?

Nana76 [90]

I believe it is, All of the above.

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3 years ago

When the distance between two objects doubles, what happens to the gravitational attraction between these two objects?

Feliz [49]

Answer:

As indicated by Newton's law of attraction each article or body in the universe draws in every single item towards one another and that power of fascination is straightforwardly relative to the result of their masses and contrarily corresponding to the square of the distance between them.

The power of gravity between two articles will diminish as the distance between them increments. The two most significant elements influencing the gravitational power between two items are their mass and the distance between their focuses. As mass increments, so does the power of gravity, however an increment in distance mirrors a reverse proportionality, which makes that power decline dramatically.

At that point by Newton's All inclusive Law of Attractive energy;

F=GMm/R^2

Mm= result of the majority

R=Distance Between the two masses by focus.

On the off chance that R is multiplied, new force=GMm/(2R)^2

=GMm/4R^2

Unique Power/New Force=4/1

F/4=New Power

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3 years ago

Two disks of polaroid are aligned so that they polarize light in the same plane. Calculate the angle through which one sheet nee

Olegator [25]

Answer: The unpolarized light's intensity is reduced by the factor of two when it passes through the polaroid and becomes linearly polarized in the plane of the Polaroid. When the polarized light passes through the polaroid with the plane of polarization at an angle $\theta$ with respect to the polarization plane of the incoming light, the light's intensity is reduced by the factor of $\cos^2\theta$ (this is the Law of Malus).

Explanation: Let us say we have a beam of unpolarized light of intensity $I_0$ that passes through two parallel Polaroid discs with the angle of $\theta$ between their planes of polarization. We are asked to find $\theta$ such that the intensity of the outgoing beam is $I_2$ . To solve this we follow the steps below:

Step 1. It is known that when the unpolarized light passes through a polaroid its intensity is reduced by the factor of two, meaning that the intensity of the beam passing through the first polaroid is

$I_1=\frac{I_0}{2}.$

This beam also becomes polarized in the plane of the first polaroid.

Step 2. Now the polarized beam hits the surface of the second polaroid whose polarization plane is at an angle $\theta$ with respect to the plane of the polarization of the beam. After passing through the polaroid, the beam remains polarized but in the plane of the second polaroid and its intensity is reduced, according to the Law of Malus, by the factor of $\cos^2\theta.$ This yields $I_2=I_1\cos^2\theta$ . Substituting from the previous step we get

$I_2=\frac{I_0}{2}\cos^2\theta$

yielding

$\frac{2I_2}{I_0}=\cos^2\theta$

and finally,

$\theta=\arccos\sqrt{\frac{2I_2}{I_0}}$

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3 years ago