3 years ago

I dont know what pfp to put on today so I'm asking u guys....

Physics

1 answer:

Mandarinka [93]3 years ago

5 0

Answer:

i think number 2 should be your pfp

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Low blood pressure. The person could faint and have an irregular heartbeat.

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3 years ago

A race car traveling at 10 meters per second accelerates at 1.5 meters per second squared while moving a distance of 600 meters.

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Answer:

Explanation:

Givens

vi = 10 m/s

a = 1.5 m/s^2

d = 600 m

vf = ?

Formula

vf^2 = vi^2 + 2*a*d

Solution

vf^2 = 10^2 + 2*1.5 * 600

vf^2 = 100 + 1800

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sqrt(vf^2) = sqrt(1900)

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A sphere of radius R contains charge Q spread uniformly throughout its volume. Find an expression for the electrostatic energy c

tensa zangetsu [6.8K]

Answer:

$E = \frac{3kQ^2}{5R}$

Explanation:

Let the sphere is uniformly charge to radius "r" and due to this charged sphere the electric potential on its surface is given as

$V = \frac{kq}{r}$

now we can say that

$q = \frac{Q}{\frac{4}{3}\pi R^3} (\frac{4}{3}\pi r^3)$

$q = \frac{Qr^3}{R^3}$

now electric potential is given as

$V = \frac{k\frac{Qr^3}{R^3}}{r}$

$V = \frac{kQr^2}{R^3}$

now work done to bring a small charge from infinite to the surface of this sphere is given as

$dW = V dq$

$dW = \frac{kQr^2}{R^3} dq$

here we know that

$dq = \frac{3Qr^2dr}{R^3}$

now the total energy of the sphere is given as

$E = \int dW$

$E = \int_0^R \frac{kQr^2}{R^3} (\frac{3Qr^2dr}{R^3})$

$E = \frac{3kQ^2}{R^6} (\frac{R^5}{5} - 0)$

$E = \frac{3kQ^2}{5R}$

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4 years ago