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-Dominant- [34]
3 years ago
13

Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of -0.9537 N when separated by

50 cm, center-to-center. The spheres are then connected by a thin conducting wire. When the wire is removed, the spheres repel each other with an electrostatic force of 0.0756 N.
What were the initial charges on the spheres?
Physics
1 answer:
scoray [572]3 years ago
8 0

Answer:

<em>The initial charges on the spheres are  </em>6.796\ 10^{-6}\ c and -3.898\ 10^{-6}\ c

Explanation:

<u>Electrostatic Force </u>

Two charges q1 and q2 separated a distance d exert a force on each other which magnitude is computed by the known Coulomb's formula

\displaystyle F=\frac{K\ q_1\ q_2}{d^2}

We are given the distance between two unknown charges d=50 cm = 0.5 m and the attractive force of -0.9537 N. This means both charges are opposite signs.

With these conditions we set the equation

\displaystyle F_1=\frac{K\ q_1\ q_2}{0.5^2}=-0.9537

Rearranging

\displaystyle q_1\ q_2=\frac{-0.9537(0.5)^2}{k}

Solving for q1.q2

\displaystyle q_1\ q_2=-2.6492.10^{-11}\ c^2\ \ ......[1]

The second part of the problem states the spheres are later connected by a conducting wire which is removed, and then, the spheres repel each other with an electrostatic force of 0.0756 N.

The conducting wire makes the charges on both spheres to balance, i.e. free electrons of the negative charge pass to the positive charge and they finally have the same charge:

\displaystyle q=\frac{q_1+q_2}{2}

Using this second condition:

\displaystyle F_2=\frac{K\ q^2}{0.5^2}=\frac{K(q_1+q_2)^2}{(4)0.5^2}=0.0756

\displaystyle q_1+q_2=2.8983\ 10^{-6}\ C

Solving for q2

\displaystyle q_2=2.8983\ 10^{-6}\ C-q_1

Replacing in [1]

\displaystyle q_1(2.8983\ 10^{-6}-q_1)=-2.64917.10^{-11}

Rearranging, we have a second-degree equation for q1.  

\displaystyle q_1^2-2.8983.10^{-6}q_1-2.64917.10^{-11}=0

Solving, we have two possible solutions

\displaystyle q_1=6,796.10^{-6}\ c

\displaystyle q_1=-3.898.10^{-6}\ c

Which yields to two solutions for q2

\displaystyle q_2=-3.898.10^{-6}\ c

\displaystyle q_2=6.796.10^{-6}\ c

Regardless of their order, the initial charges on the spheres are 6.796\ 10^{-6}\ c and -3.898\ 10^{-6}\ c

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4 0
4 years ago
Raindrops fall vertically at 7.5 m/s relative to the Earth. What does an observer in a car moving at 20.2 m/s in a straight line
Vilka [71]

Answer:

vDP = 21.7454 m/s

θ = 200.3693°

Explanation:

Given

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vPE = 20.2 m/s

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vDP = - 7.5 j - 20.2 i

The magnitude of vDP is given by

vDP = √((- 7.5)²+(- 20.2)²) m/s =  21.7454 m/s

θ = Arctan (- 7.5/- 20.2) = 20.3693°

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4 0
4 years ago
A 45 kg bear slides, from rest, 11 m down a lodgepole pine tree, moving with a speed of 5.8 m/s just before hitting the ground.
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Answer:

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b )

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c ) If  the average frictional force that acts on the sliding bear be F

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Answer:

The heat transferred through the wall that day is  13728 BTUs

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Here, we have the area of the wall given as

Area of wall = 2 × Length × Height + 2 × Width × Height

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Height = 9 feet

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Temperature difference is given by

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