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miss Akunina [59]

2 years ago

7

Consider a system consisting of two Einstein solids, A and B, each containing N= 10 oscillators, and sharing a total of q= 20 un

its of energy. Assume that the two solids areweakly coupled and the total energy is fixed.
Required:
a. How many different macrostates are available to solid A or B?
b. How many different microstates are available to the combined system of A & B?
c. Starting with the exact form for the multiplicity of an Einstein solid, calculate the probability of finding all the energy in solid A, assuming that the system is in thermal equilibrium.
d. Calculate the probability of finding exactly half the energy in solid A.

1 answer:

Daniel [21]2 years ago

7 0

Answer:Gg

Explanation:

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William Ferrel:

balu736 [363]

Answer:

William Ferrel created a tide-prediction machine.

Explanation:

William Ferrel create a machine in late 19th century that was the best combination of mechanical parts and computer coding.
It was a mechanical analog computer that could predict the ebb of tides and even the height of tides that could be irregular.
It was widely used for marine networks and navigation. Later on many improvisations and additional features were added on it.
During the world war times, this tide prediction machine was of great use for military purpose.

3 0

3 years ago

Which of these is true about kinetic energy but not necessarily true about potential energy

cestrela7 [59]

Kinetic energy is never negative, but potential energy can be.

Potential energy depends on height above some reference level,
and you can pick any level you want as the reference. So, if the
object is below the reference level you pick, then its potential
energy relative to your reference level is negative.

What that means is: You have to lift it / do work on it / give it more
energy than it has now ... in order to move it to the reference level.

(That's exactly the situation with electrons bound to an atom. Their
energy is considered negative, because we have to do work and
give them more energy to rip them away from the atom.)
_____________________________________

Regarding the other choices:

-- Kinetic energy is scalar ... Yes. So is potential energy.

-- Kinetic energy increases with height ...
No. It doesn't, but potential energy does.

-- Kinetic energy depends on position ...
No. It doesn't, but potential energy does.

3 0

3 years ago

In a local bar, a customer slides an empty beer mug down the counter for a refill. The height of the counter is 1.42 m. The mug

telo118 [61]

Answer:

a) $V_{x}=3.72m/s$ , b) ∠=-54.83°

Explanation:

In order to solve this problem, we must start with a drawing of the situation, this will help us visualize the problem better. (See picture attached).

a)

Now, the idea is that the beer mug has a horizontal speed and no vertical speed at initial conditions. So knowing this, we can start finding the initial velocity of the mug.

In order to do so, we need to find the time it takes for the mug to reach the ground. We can find it by using the following equation:

$y=y_{0}+V_{y0}t+\frac{1}{2}a_{y}t^{2}$

We can see from the drawing that y and the initial velocity in y are zero, so we can simplify our formula:

$0=y_{0}+\frac{1}{2}a_{y}t^{2}$

so we can solve for t, so we get:

$t=\sqrt{\frac{-(2)y_{0}}{a}}$

so now we can substitute the known values, so we get:

$t=\sqrt{\frac{-(2)(1.42)}{-9.8}}$

which yields:

t=0.538s

So we can use this value to find the velocity in x:

$V_{x}=\frac{x}{t}$

When substituting we get:

$V_{x}=\frac{2m}{0.538s}$

which yields:

$V_{x}=3.72m/s$

b)

In order to solve part b, we need to find the y-component of the velocity, for which we can use the following formula:

$\Delta y=\frac{V_{f}^{2}-V_{0}^{2}}{2a}$

We know that $V_{0}$ is zero, so we can simplify the expression:

$\Delta y=\frac{V_{yf}^{2}}{2a}$

So we can solve the equation for $V_{yf}^{2}$ so we get:

$V_{yf}=\sqrt{2\Delta y a}$

and when substituting the known values we get:

$V_{yf}=\sqrt{2(-1.42m)(-9.8m/s^{2})}$

which yields:

$V_{yf}=-5.28m/s$

Once we got the final velocity in y, we can use it together with the velocity in x to find the angle.

So we can use the following formula:

$tan \theta =\frac{V_{y}}{V_{x}}$

when solving for theta we get:

$\theta = tan^{-1}(\frac{V_{y}}{V_{x}})$

We can substitute so we get:

$\theta = tan^{-1}(\frac{-5.28m/s}{3.72m/s})$

which yields:

$\theta = -54.83^{o}$

7 0

3 years ago

The load across a 12 v battery consists of a series combination of three resistors 45 ω, 55 ω, and 35 ω. what is the total resis

zubka84 [21]

For n resistors in series, the equivalent resistance is given by the sum of the resistances:
$R_{eq} = R_1 + R_2 + ... + R_n$

In this problem, we have three resistors, so the equivalent resistance of the load is the sum of the resistances of the three resistors:
$R_{eq}=45 \Omega + 55 \Omega + 35 \Omega =135 \Omega$

5 0

3 years ago

An electric turntable 0.730 mm in diameter is rotating about a fixed axis with an initial angular velocity of 0.240 rev/srev/s a

Zolol [24]

Answer:

a) $\omega = 0.421\,\frac{rev}{s}$ , b) $\Delta \theta = 0.066\,rev$ , c) $v = 0.966\,\frac{mm}{s}$ , d) $a = 3.293\,\frac{mm}{s^{2}}$

Explanation:

a) The angular velocity of the turntable after $0.200\,s$ .

$\omega = \omega_{o} + \alpha\cdot \Delta t$

$\omega = 0.240\,\frac{rev}{s} + (0.906\,\frac{rev}{s^{2}} )\cdot (0.2\,s)$

$\omega = 0.421\,\frac{rev}{s}$

b) The change in angular position is:

$\Delta \theta = \omega_{o}\cdot t + \frac{1}{2} \cdot \alpha \cdot t^{2}$

$\Delta \theta = (0.240\,\frac{rev}{s} )\cdot (0.2\,s) + \frac{1}{2}\cdot (0.906\,\frac{rev}{s^{2}} )\cdot (0.2\,s)^{2}$

$\Delta \theta = 0.066\,rev$

c) The tangential speed of a point on the rim of the turn-table:

$v = r\cdot \omega$

$v = (0.365\times 10^{-3}\,m)\cdot (0.421\,\frac{rev}{s} )\cdot (\frac{2\pi\,rad}{1\,rev} )$

$v = 9.655\times 10^{-4}\,\frac{m}{s}$

$v = 0.966\,\frac{mm}{s}$

d) The tangential and normal components of the acceleration of the turn-table:

$a_{t} = (0.365\times 10^{-3}\,m)\cdot (0.906\,\frac{rev}{s^{2}})\cdot (\frac{2\pi\,rad}{1\,rev} )$

$a_{t} = 2.078\times 10^{-3}\,\frac{m}{s^{2}}$

$a_{t} = 2.078\,\frac{mm}{s}$

$a_{n} = (0.365\times 10^{-3}\,m)\cdot \left[(0.421\,\frac{rev}{s} )\cdot (\frac{2\pi\,rad}{1\,rev} )\right]^{2}$

$a_{n} = 2.554\times 10^{-3}\,\frac{m}{s^{2}}$

$a_{n} = 2.554\,\frac{mm}{s^{2}}$

The magnitude of the resultant acceleration is:

$a = \sqrt{(2.078\,\frac{mm}{s} )^{2}+(2.554\,\frac{mm}{s} )^{2}}$

$a = 3.293\,\frac{mm}{s^{2}}$

8 0

3 years ago