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belka [17]
3 years ago
11

Two identical point charges are fixed to diagonally opposite corners of a square that is 0.510 m on a side. Each charge is 3.03

10-6 C. How much work is done by the electric force as one of the charges moves to an empty corner?
Physics
1 answer:
Evgen [1.6K]3 years ago
3 0

Answer:

Workdone = 0.05Joules

Explanation:

The side of a square =a= 0.510m

q1=q2=3.03×10^-6C

W=UF-Ui

W= kq1q2/a - kq1q2/r

But r= sqrt(a^2+a^2)=sqrt2a

W=kq1q2/a - kq1q1/sqrt2a

W= kq1q2/a (1 -1/sqrt2)

W= (9×10^9)(3.03×10^-6)(3.03×10^-6)(1-1/sqrt2)

W= 0.05Joules

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Answer:

Except in the crust, the interior of the Earth cannot be studied by drilling holes to take samples. Instead, scientists map the interior by watching how seismic waves from earthquakes are bent, reflected, sped up, or delayed by the various layers.

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Ok, now what do you want to know about it?
4 0
3 years ago
How many excess electrons must be present on each sphere if the magnitude of the force of repulsion between them is 3.33××10âˆ
AlekseyPX

Answer:

There are 756.25 electrons present on each sphere.

Explanation:

Given that,

The force of repression between electrons, F=3.33\times 10^{-21}\ N

Let the distance between charges, d = 0.2 m

The electric force of repulsion between the electrons is given by :

F=k\dfrac{q^2}{r^2}

q=\sqrt{\dfrac{Fr^2}{k}}

q=\sqrt{\dfrac{3.33\times 10^{-21}\times (0.2)^2}{9\times 10^9}}

q=1.21\times 10^{-16}\ C

Let n are the number of excess electrons present on each sphere. It can be calculated using quantization of charges. It is given by :

q = ne

n=\dfrac{q}{e}

n=\dfrac{1.21\times 10^{-16}}{1.6\times 10^{-19}}

n = 756.25 electrons

So, there are 756.25 electrons present on each sphere. Hence, this is the required solution.

8 0
3 years ago
A box is placed on a 30o frictionless incline. What is the acceleration of the box as it slides down the incline
balandron [24]

Answer:

<em>2.78m/s²</em>

Explanation:

Complete question:

<em>A box is placed on a 30° frictionless incline. What is the acceleration of the box as it slides down the incline when the co-efficient of friction is 0.25?</em>

According to Newton's second law of motion:

\sum F_x = ma_x\\F_m - F_f = ma_x\\mgsin\theta - \mu mg cos\theta = ma_x\\gsin\theta - \mu g cos\theta = a_x\\

Where:

\mu is the coefficient of friction

g is the acceleration due to gravity

Fm is the moving force acting on the body

Ff is the frictional force

m is the mass of the box

a is the acceleration'

Given

\theta = 30^0\\\mu = 0.25\\g = 9.8m/s^2

Required

acceleration of the box

Substitute the given parameters into the resulting expression above:

Recall that:

gsin\theta - \mu g cos\theta = a_x\\

9.8sin30 - 0.25(9.8)cos30 = ax

9.8(0.5) - 0.25(9.8)(0.866) = ax

4.9 - 2.1217 = ax

ax = 2.78m/s²

<em>Hence the acceleration of the box as it slides down the incline is 2.78m/s²</em>

5 0
3 years ago
A linear network has a current input 7.5 cos(10t + 30°) A and a voltage output 120 cos(10t + 75°) V. Determine the associated im
Leona [35]

Answer:

16∠45° Ω

Explanation:

Applying,

Z = V/I................... Equation 1

Where Z = Impedance, V = Voltage output, I = current input.

Given: V = 120cos(10t+75°), = 120∠75°,  I = 7.5cos(10t+30) = 7.5∠30°

Substitute these values into equation 1

Z = 120cos(10t+75°)/7.5cos(10t+30)

Z = 120∠75°/ 7.5∠30°

Z = 16∠(75°-30)

Z = 16∠45° Ω

Hence the impedance of the linear network is 16∠45° Ω

8 0
3 years ago
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