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Ganezh [65]
2 years ago
9

Two small, positively charged spheres have a combined charge of 5.0 x 10 -5 C. If each sphere is repelled from the other by an e

lectrostatic force of 1.0 N when the spheres are 2.0 m apart, what is the charge, in micro-coulomb, on the sphere with the smaller charge?
Physics
1 answer:
geniusboy [140]2 years ago
4 0

Explanation:

It is given that,

Let q₁ and q₂ are two small positively charged spheres such that,

q_1+q_2=5\times 10^{-5}\ C.............(1)

Force of repulsion between the spheres, F = 1 N

Distance between spheres, d = 2 m

We need to find the charge on the sphere with the smaller charge. The force is given by :

F=k\dfrac{q_1q_2}{d^2}

q_1q_2=\dfrac{F.d^2}{k}

q_1q_2=\dfrac{1\ N\times (2\ m)^2}{9\times 10^9}

q_1q_2=4.45\times 10^{-10}\ C............(2)

On solving the system of equation (1) and (2) using graph we get,

q_1=0.0000384\ C=38.4\ \mu C

q_2=0.0000116\ C=11.6\ \mu C

So, the charge on the smaller sphere is 11.6 micro coulombs. Hence, this is the required solution.

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tia_tia [17]

Answer:

Explanation:

a ) Earlier emf of cell applied on R₁ but now emf will be distributed among R₁ and R₂

Potential difference on R₁ will become less .

b ) Current is inversely proportional to resistance of the circuit. As resistance increases , current will be less . So current through R₁ will become less.

c )

When resistance is added in series , they are added up to obtain equivalent resistance . So equivalent resistance R₁₂ will be more than R₁ OR R₂.

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What's a difference between mercury and oxygen?<br> (They both are elements)
borishaifa [10]

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3 years ago
How much does the earth/sky weigh?
Talja [164]
Calculated weight (by experimentally) of Earth is 5.972 × 10²⁴ N

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6 0
2 years ago
Two people carry a heavy object by placing it on a board that is 2.2 meters long. One person lifts with a force of 750 N at one
stiks02 [169]

Answer:

This means that the center of mass is locates 0.72m from the 750N force

Explanation:

Since the board is 2.2m long, that will be the length of the board.

Let the center of mass of the body be hinged at the center using a knife edge as shown in the diagram attached.

Let x be the distance from the 750N force to the knife edge and the distance from the 360N force to the knife edge be 2.2-x

Using the principle of moment which states that the sum of clockwise moment is equal to the sum of anti clockwise moment.

Moment = force × perpendicular distance

For ACW moment;

Moment = 750×x = 750x

For the CW moment;

Moment = 360 × (2.2-x)

Moment = 792-360x

Equating ACW moment to the clockwise moment we have;

750x = 792-360x

750x+360x = 792

1110x = 792

x = 792/1110

x = 0.72m

This means that the center of mass is locates 0.72m from the 750N force

3 0
2 years ago
A phone cord is 4.89 m long. The cord has a mass of 0.212 kg. A transverse wave pulse is produced by plucking one end of the tau
slega [8]

Answer:

Tension in Cord=174 N

Explanation:

Given Data

L (Phone Cord Length)=4.89 m

m (Cord Mass)=0.212 Kg

T (Time for four trips)=0.617 s

Tension=?

Solution

V=λ×f

V=\frac{8*4.89}{0.617}\\ V=63.4m/s

Sigma=\frac{mass}{length}\\ Sigma=\frac{0.212}{4.89}\\ Sigma=0.0433 \frac{kg}{m}

Wave Speed=\sqrt{\frac{Tension}{Sigma} }\\ \\V=\sqrt{\frac{T}{Sigma} }\\ V^{2}=\frac{T}{Sigma}\\  T=V^{2}*Sigma\\ T=(63.4)^{2}*(0.0433)\\ T=174 N

5 0
2 years ago
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