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3 years ago

5

The electric potential at a certain location from a point charge can be represented by V. What is the value of the electric pote

ntial at the same location if the strength of the charge is tripled?

1 answer:

adoni [48]3 years ago

5 0

<h2>Answer:</h2>

If you triple the charge, the electric potential is 3V.

<h2>Explanation:</h2>

If you triple the charge, the electric potential is 3V. Put another way, the value of the electric potential at the same location is tripled if the strength of the charge is tripled.

<em>But why this is true?</em>

Well, the potential V due to a single point charge q is:

$V=\frac{1}{4\pi\varepsilon_{0}}\frac{q}{r}$

If the strength of the charge is tripled means that our new q1=3q, therefore:

$V=\frac{1}{4\pi\varepsilon_{0}}\frac{3q}{r}$

$V=3\frac{1}{4\pi\varepsilon_{0}}\frac{q}{r}$

So you can see that in fact if you triple the charge, the electric potential is 3V.

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mass of racket, M = 331 g = 0.331 Kg

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a) momentum of ball before collision

P₁ = m v

P₁ = 0.0575 x 26.7

P₁ = 1.535 kg.m/s

b) momentum of ball after collision

P₂ = m v'

P₂ = 0.0575 x (-29.5)

P₂ = -1.696 kg.m/s

c) change in momentum

Δ P = P₂ - P₁

Δ P = -1.696 -1.535

Δ P = -3.231 kg.m/s

d) using conservation of momentum

initial speed of racket = 0 m/s

M u + m v = Mu' + m v

M x 0 + 0.0575 x 26.7 = 0.331 x u' + 0.0575 x (-29.5)

0.331 u' = 3.232

u' = 9.76 m/s

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3 years ago

A wire of radius R has a current I uniformly distributed across its cross-sectional area. Ampere's law is used with a concentric

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Answer:

Please refer to the figure.

Explanation:

The crucial point here is to calculate the enclosed current. If the current I is flowing through the whole cross-sectional area of the wire, the current density is

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The current density is constant for different parts of the wire. This idea is similar to that of the density of a glass of water is equal to the density of a whole bucket of water.

So,

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This enclosed current is now to be used in Ampere’s Law.

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Here, $\int \, dl$ represents the circular path of radius r. So we can replace the integral with the circumference of the path, $2\pi r$ .

As a result, the magnetic field is

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2 years ago

A flywheel is a solid disk that rotates about an axis that is perpendicular to the disk at its center. Rotating flywheels provid

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Answer:

$8.37*10^5$ rpm

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t3 = 2 min = 2 x 60 = 120 s

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