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4 years ago

A particle with charge 8 µC is located on the

Physics

1 answer:

Oksi-84 [34.3K]4 years ago

8 0

Answer: 34.65 N towards charge 8 μC.

Explanation:

The electrostatic force between two charges is given by:

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

where, k is the Coulomb constant = 8.9875 × 10⁹ N.m²/C²

q₁ and q₂ are the two charges separated by distance r.

The distance between charges 8 μC and -7 μC is r = 2 cm -(-10 cm) = 12 cm = 0.12 m

The force between these charges is:

$F = 8.9875 \times 10^9 Nm^2/C^2 \times \frac{8\times 10^{-6}C\times (-7\times10^{-6}C)}{(0.12m)^2} = -34.95 N$

Negative sign implies it is an attractive force.

The distance between -3 μC charge and -7 μC charge is r' = 10 cm -2 cm = 8 cm = 0.8 m.

The electrostatic force between these charges is:

$F' = 8.9875\times 10^9Nm^2/C^2\times \frac{(-3 \times 10^{-6}C)\times (-7\times 10^{-6}C)}{(0.8m)^2} = 0.295 N$

It is a repulsive force.

Net force on the -7 μC charge is:

Fn = F + F'

we can add them directly as they are acting in one direction along the x-axis.

Fn = -34.95 N + 0.295 = -34.65 N

Thus, the net force is attractive in nature and it is towards charge 8 μC.

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Consider a positive charge Q and a point B twice as far away from Q as point A. What is the ratio of the electric field strength

Vikentia [17]

Answer:

$\frac{E_{A}}{E_{B}}=4$

Explanation:

The electric field is defined as the electric force per unit of charge, this is:

$E=\frac{F}{q}$ .

The electric force can be obtained through Coulomb's law, which states that the electric force between to electrically charged particles is inversely proportional to the square of the distance between them and directly proportional to the product of their charges. The electric force can be expressed as

$F=\frac{kQq}{r^{2}}$ .

By substitution we get that

$E=\frac{kQq}{qr^{2}}\\\\E=\frac{kQ}{r^{2}}$

Now, letting $E_{A}$ be the electric field at point A, letting $E_{B}$ be the electric field at point B, and letting R be the distance from the charge to A:

$E_{A}=\frac{kQ}{R^{2}}\\\\E_{B}=\frac{kQ}{(2R)^{2}}$ .

The ration of the electric fields is

$\frac{E_{A}}{E_{B}}=\frac{\frac{kQ}{R^{2}}}{\frac{kQ}{(2R)^{2}}}\\\\\frac{E_{A}}{E_{B}}=\frac{\frac{1}{R^{2}}}{\frac{1}{(2R)^{2}}}\\\\\frac{E_{A}}{E_{B}}=\frac{\frac{1}{R^{2}}}{\frac{1}{(4)R^{2}}}\\\\\\\frac{E_{A}}{E_{B}}=\frac{1}{\frac{1}{(4)}}\\\\\frac{E_{A}}{E_{B}}=4$

This means that at half the distance, the electric field is four times stronger.

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3 years ago

Whenever a musician plays a guitar, they pluck one of the guitar strings to produce a standing wave in the string. Then pinch do

UkoKoshka [18]

Answer:

It's due to the distance from either ends of strings origin...

Explanation:

As we know that waves behave moving in a flow from one side to another side and this gives a prospective of motion. Suppose a wave is pinched from the near one end of a guitar then due to the distortion created by the point of tie of strings the wave super imposes and moves with a velocity v and produces a wave frequency f. as we the pinching go down to the center the wave stabilizes itself to a stationary origin right at the center and the frequency then changes accordingly as moving down on the string.

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3 years ago

A girl (mass M) standing on the edge of a frictionless merry-go-round (radius R, rotational inertia I) that is not moving. She t

vladimir1956 [14]

a) $\omega=\frac{-mvR}{I+MR^2}$

b) $v=\frac{-mvR^2}{I+MR^2}$

Explanation:

Since there are no external torques acting on the system, the total angular momentum must remain constant.

At the beginning, the merry-go-round and the girl are at rest, so the initial angular momentum is zero:

$L_1=0$

Later, after the girl throws the rock, the angular momentum will be:

$L_2=(I_M+I_g)\omega +L_r$

where:

$I$ is the moment of inertia of the merry-go-round

$I_g=MR^2$ is the moment of inertia of the girl, where

M is the mass of the girl

R is the distance of the girl from the axis of rotation

$\omega$ is the angular speed of the merry-go-round and the girl

$L_r=mvR$ is the angular momentum of the rock, where

m is the mass of the rock

v is its velocity

Since the total angular momentum is conserved,

$L_1=L_2$

So we find:

$0=(I+I_g)\omega +mvR\\\omega=\frac{-mvR}{I+MR^2}$

And the negative sign indicates that the disk rotates in the direction opposite to the motion of the rock.

The linear speed of a body in rotational motion is given by

$v=\omega r$

where

$\omega$ is the angular speed

r is the distance of the body from the axis of rotation

In this problem, for the girl, we have:

$\omega=\frac{-mvR}{I+MR^2}$ is the angular speed

$r=R$ is the distance of the girl from the axis of rotation

Therefore, her linear speed is:

$v=\omega R=\frac{-mvR^2}{I+MR^2}$

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3 years ago

An airplane has wings, each with area A, designed so that air flows over the top of the wing at 265 m/s and underneath the wing

exis [7]

Answer

given,

Pressure on the top wing = 265 m/s

speed of underneath wings = 234 m/s

mass of the airplane = 7.2 × 10³ kg

density of air = 1.29 kg/m³

using Bernoulli's equation

$P_1 + \dfrac{1}{2}\rho v_1^2 = P_2 + \dfrac{1}{2}\rho v_2^2$

$\Delta P =\dfrac{1}{2}\rho (v_2^2-v_1^2)$

$\Delta P =\dfrac{1}{2}\times 1.29\times (265^2-234^2)$

$\Delta P =9977.5 Pa$

Applying newtons second law

2 Δ P x A - mg = 0

$A =\dfrac{mg}{2\Delta P}$

$A =\dfrac{7.2\times 10^3 \times 9.8}{2\times 9977.5}$

A = 3.53 m²

7 0

3 years ago

a refridgerator has a coefficient of performance equal to 4.2 how much work must be done on the refridgerator in order to remove

anastassius [24]

Answer:

60 J

Explanation:

Given That

Coefficient of performance, CoP = 4.2

Quantity of heat of the refrigerator, Q = 250 J

Work done by the refrigerator, W = ?

First, it should be noted that the Coefficient of Performance can be said to be, the ratio of the Heat required to the Work done by the system. Mathematically written as,

CoP = Q / W

Since we are already given the values from our question, we can plug it in as it's a pretty straightforward question

4.2 = 250 / W, making W subject of formula, we have

W = 250 / 4.2

W = 60J

Thus, the Work done by the refrigerator is 60 J

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3 years ago