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3 years ago

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Consider the Bose-Einstein distribution: F60 per unit energy can be written as N aE 6.) (20pts) Assuming BE <<1,obtain an

expression for the average number of particles in the energy range 0SESE 7) (10pts) Assuming T-300.0K, E -3.5 x 102 J, α = (3.8 x 103 ) /ji , what İs the average number of particles? Suppose the number of state

1 answer:

notsponge [240]3 years ago

4 0

Answer:

Explanation:

Whacbdhnd

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A 20 cm-radius ball is uniformly charged to 71 nC.

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Answer:

Part a)

$\rho = 2.12\mu C/m^3$

Part b)

$q_1 = 1.11 nC$

$q_2 = 8.88 nC$

$q_3 = 71 nC$

Part c)

$E_1 = 3996 N/C$

$E_2 = 7992 N/C$

$E_3 = 15975 N/C$

Explanation:

Part a)

As we know that charge density is the ratio of total charge and total volume

So here the volume of the charge ball is given as

$V = \frac{4}{3}\pi R^3$

$V = \frac{4}{3}\pi(0.20)^3$

$V = 0.0335 m^3$

now the charge density of the ball is given as

$\rho = \frac{71 nC}{0.0335} = 2.12\mu C/m^3$

Part b)

Now the charge enclosed by the surface is given as

$q = \rho V$

at radius of 5 cm

$q = (2.12 \mu C/m^3)(\frac{4}{3}\pi(0.05)^3$

$q = 1.11 nC$

at radius of 10 cm

$q = (2.12 \mu C/m^3)(\frac{4}{3}\pi(0.10)^3$

$q = 8.88 nC$

at radius of 20 cm

$q = 71 nC$

Part c)

As we know that electric field is given as

$E = \frac{kq}{r^2}$

so we have electric field at r = 5 cm

$E_1 = \frac{(9\times 10^9)(1.11 nC)}{0.05^2}$

$E_1 = 3996 N/C$

electric field at r = 10 cm

$E_2 = \frac{(9\times 10^9)(8.88 nC)}{0.10^2}$

$E_2 = 7992 N/C$

electric field at r = 20 cm

$E_3 = \frac{(9\times 10^9)(71 nC)}{0.20^2}$

$E_3 = 15975 N/C$

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2 years ago

Toy car in a science experiment covers 1.6 meters in half a second. If a the car travels at a steady speed, how far will it go i

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The answer is D. 32 m.

The simple equation that connects speed (v), time (t), and distance (d) can be expressed as:
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d = ?

So:
$d= v*t=3.2m/s*10s = 32m$

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