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3 years ago

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For a two-level system, the weight of a given energy distribution can be expressed in terms of the number of systems, N, and the

number of systems occupying the excited state, n1. What is the expression for weight in terms of these quantities

1 answer:

Nikitich [7]3 years ago

7 0

Answer:

W = N!/(n0! * n1!)

Explanation:

Let n0 = number of particles in the lowest energy state

n1 = number of particles in the excited energy state.

Using this, we can say that N = n0 + n1

From this we can then express the weight, W of the close system by finding the factorials of each particles

W = N!/(n0! * n1!)

Hence, the weight W is expressed as W = N!/(n0! * n1!)

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A 200 kg weather rocket is loaded with 100 kg of fuel and fired straight up. It accelerates upward at 30 m/s2 for 35 s , then ru

Gre4nikov [31]

Answer:

a) $h_m=74625\ m$

b) $t=265.55\ s$

Explanation:

Given:

mass of rocket, $m_r=200\ kg$

mass of fuel, $m_f=100\ kg$

acceleration of the rocket consuming fuel, $a=30\ m.s^{-2}$

time after which the fuel exhaust, $t_f=35\ s$

<u>During the phase of fuel exhaustion:</u>

<u>velocity attained by the rocket just as the fuel ends:</u>

$v_f=u+a.t_f$

where:

$u=$ initial velocity of the rocket = 0

$v_f=0+30\times 35$

$v_f=1050\ m.s^{-1}$ this will be the initial velocity for the phase of ascending of the rocket's height under the influence of gravity.

<u>height at which the fuel finishes:</u>

$v_f^2=u^2+2a.h_f$

$1050^2=0^2+2\times 30\times h_f$

$h_f=18375\ m$

<u>During the phase of ascend in height of rocket after the fuel is over:</u>

<u>Time taken to reach the top height after the fuel is over:</u>

$v=v_f+g.t'$

at top v = (final velocity during this course of motion )= 0 $m.s^{-1}$

$0=1050-9.8\times t'$

$t'=107.1429\ s$

<u>Height ascended by the rocket after the fuel is over:</u>

$v^2=v_f^2+2g.h'$

at the top height the velocity is zero

$0^2=1050^2-2\times 9.8\times h'$ (-ve sign denotes that the direction of motion is opposite to that of acceleration)

$h'=56250\ m$

<u>Therefore the maximum altitude attained by the rocket:</u>

$h_m=h_f+h'$

$h_m=18375+56250$

$h_m=74625\ m$

b)

time taken by the rocket to fall back to the earth:

$h_m=v.t_m+\frac{1}{2} g.t_m^2$

where:

$v=$ initial velocity of the rocket during the course of free fall from the top height.

$74625=0+4.9\times t_m^2$

$t_m=123.41\ s$

Now the total time for which the rocket is in the air:

$t=t_f+t'+t_m$

$t=35+107.1429+123.41$

$t=265.55\ s$

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3 years ago

What area of earth contains semi-solid rock and lava?

ioda

Below the crust is the mantle, a dense, hot layer of semi-solid rock approximately 2,900 km thick. The mantle, which contains more iron, magnesium, and calcium than the crust, is hotter and denser because temperature and pressure inside the Earth increase with depth.

Hope this helped! :)
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3 years ago

A 360-nm thick oil film floats on the surface of a pool of water. the indices of refraction of the oil and the water are 1.5 and

jenyasd209 [6]

The correct answer is 432, and 720.
The thickness of a film is t= 360nm
the refractive index of oil n₀t = (m +1/2) λ
For m =0
λ = 4n₀t
= 4(1.50)(360)
= 2160nm
for m = 1
λ = 4n₀t
= 4(1.50)(360)/3
= 720nm
m = 2
λ = 4n₀t/5 = 4(1.50)(360)/5
= 432nm
The wavelength which are most strongly reflected are
432nm, 720nm.

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3 years ago

While running these tests, crall notices a similarity in the velocity measured at the ground independent of whether the tennis b

lara [203]

At the ground the ball will always have velocity along the direction of gravity. If upward motion is taken positive it will always have negative velocity at the ground because, if the ball was given an initial upward velocity then gravity will decelerate it and bring it down with a negative final velocity. If the ball is given an initial downward velocity then the ball will be further accelerated by gravity in the downward direction only, again maintaining negative direction. The magnitude however in both cases will be different. the final velocity at the ground will have higher magnitude in case of elevator moving downwards.

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3 years ago

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2. Adelia holds a shiny steel spoon with its back (convex surface) facing her eyes at a distance

My name is Ann [436]

Answer:

(a) The convex mirror image, is always upright at all positions, while images formed by concave mirrors are always inverted when the object distance from the mirror is more than the mirrors focal length.

(b) An upright image is not seen for object at a distance from a concave mirror further than the focal length of the mirror, which is the spoon in the question

Therefore, the location of her eyes of approximately, 30 cm, from the mirror is more than the mirror's focal length

Explanation:

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3 years ago