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2 years ago

7

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]2 years ago

4 0

Answer:

Explanation:

Whacbdhnd

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3. The electric field of a sinusoidal electromagnetic wave has an amplitude of 5.0 V/m. How much radiation energy passes through

True [87]

Answer:

e) $179$ J

Explanation:

$E_{o}$ = Magnitude of electric field = 5 V/m

$A$ = Area of window = 1.5 m²

$c$ = speed of electromagnetic wave = 3 x 10⁸ m/s

$\Delta t$ = time interval = 1 h = 3600 sec

radiation energy is given as

$U = (0.5)\epsilon _{o}E_{o}^{2}cA\Delta t$

$U = (0.5)(8.85\times 10^{-12})(5)^{2}(3\times 10^{8})(1.5)(3600)$

$U = 179$ J

6 0

2 years ago

A 12 kg block is accelerating at the rate of 9.0 m/s? while being acted on by two forces.

Aliun [14]

Answer:

See below

Explanation:

F = ma

F = 12 * 9 = 108 N

108 N needed <u> add 30 N more east </u>

8 0

9 months ago

A metal container with the oil weigh

MA_775_DIABLO [31]

Answer:

16 kg

Explanation:

M - container

m - oil mass

by definition of density ,

relative density is the ratio of the density of a substance to the density of water.

relative density = density/ density of water

density of oil = 1.2*1000 kgm⁻³ = 1200 kgm⁻³

1 Litre =10⁻³ m³

oil volume = 80 *10⁻³ m³

mass of oil = density * volume

= 1200*80*10⁻³

= 96 kg

Mass of container + mass of oil =112

mass of container = 112 - 96

= 16 kg

7 0

2 years ago

6.If a 250. gram cart moving to the right with a velocity of .31 m/s collides inelastically with a 500. gram cart traveling to t

spin [16.1K]

Answer:

The total momentum of the system before the collision is 0.0325 kg-m/s due east direction.

Explanation:

Given that,

Mass of the cart, m = 250 g = 0.25 kg

Initial velocity of the cart, u = 0.31 m/s (due right)

Mass of another cart, m' = 500 g = 0.5 kg

Initial velocity of the another cart u' = -0.22 m/s (due left)

Let p is the total momentum of the system before the collision. It is given by :

$p=mu+m'u'\\\\p=0.25\times 0.31+0.5\times (-0.22)\\\\p=-0.0325\ kg-m/s$

So, the total momentum of the system before the collision is 0.0325 kg-m/s due east direction.

5 0

2 years ago

Read 2 more answers

A thermometer is removed from a room where the temperature is 70° F and is taken outside, where the air temperature is 10° F. Af

vekshin1

Answer:

$T=51.64^\circ F$

$t=180.10s$

Explanation:

The Newton's law in this case is:

$T(t)=T_m+Ce^{kt}$

Here, $T_m$ is the air temperture, C and k are constants.

We have

$70^\circ F$ in $t=0$

So:

$T(0)=70^\circ F\\T(0)=10^\circ F+Ce^{k(0)}\\70^\circ F=10^\circ F+C\\C=70^\circ F-10^\circ F=60^\circ F$

And we have $60^\circ F$ in $t=30 s$ , So:

$T(30)=60^\circ F\\T(30)=10^\circ F+(60^\circ F)e^{k(30)}\\60^\circ F=10^\circ F+(60^\circ F)e^{k(30)}\\50^\circ F=(60^\circ F)e^{k(30)}\\e^{k(30)}=\frac{50^\circ F}{60^\circ F}\\(30)k=ln(\frac{50}{60})\\k=\frac{ln(\frac{50}{60})}{30}=-0.0061$

Now, we have:

$T=10^\circ F+(60^\circ F)e^{-0.0061t}(1)$

Applying (1) for $t=1 min=60s$ :

$T=10^\circ F+(60^\circ F)e^{-0.0061*60}\\T=10^\circ F+(60^\circ F)0.694\\T=10^\circ F+41.64^\circ F\\T=51.64^\circ F$

Applying (1) for $T=30^\circ F$ :

$30^\circ F=10^\circ F+(60^\circ F)e^{-0.0061t}\\30^\circ F-10^\circ F=(60^\circ F)e^{-0.0061t}\\-0.0061t=ln(\frac{20}{60})\\t=\frac{ln(\frac{20}{60})}{-0.0061}=180.10s$

8 0

3 years ago