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baherus [9]
3 years ago
13

Suppose you had 10 identical molecules enclosed by a box. At a given instant, one molecule has an energy of 100 Joules, and the

others are all stationary
(A) What is the average kinetic energy of the 10 molecules?
(B) Is this a situation of high or low entropy?
(C) What happens to the energy and entropy of the molecules in the box as time passes? Does your answer depend on the insulation of the box?
Physics
1 answer:
boyakko [2]3 years ago
4 0

Answer:

A)   K_average = 1/20 m v², B) low entropy , C) entropy increases.

Explanation:

A) The kinetic energy of each molecule

         K = ½ m v²

The average kinetic energy is the sum of each kinetic energy among the number of them

        K_average = (½ m v² + 0 + 0 +) 10

        K_average = 1/20 m v²

B) Entropy is the sum of the states of each molecule, in this configuration there are only two states one with energy and the other with zero energy, so it is a system with low entropy.

       S = k ln W

  Where S is the entropy, k the Bolztmann constant, W the amount of state present in the system in this case is 2

C) Let's start by analyzing the entropy, as time goes by the molecule that is moving collides with the other molecules and transfers them some energy, so the other molecules move to a different state, after a little In time all the molecules will have the same energy, each one in a different state or volume, so the number of possible state increases to 10, so the entropy increases.

Now let's analyze what happens with the energy of the system, for this case we have two possibilities

- The system is isolated, therefore as it cannot exchange energy with the environment, the total energy remains constant even when the energy of each molecule can fluctuate.

- If the system is not isolated, it can exchange energy with the environment, therefore the total energy changes, depending on the difference in energy between the molecules and the environment

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4 This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive
zubka84 [21]

Answer:at 21.6 min they were separated by 12 km

Explanation:

We can consider the next diagram

B2------15km/h------->Dock

|

|

B1 at 20km/h

|

|

V

So by the time B1 leaves, being B2 traveling at constant 15km/h and getting to the dock one hour later means it was at 15km from the dock, the other boat, B1 is at a distance at a given time, considering constant speed of 20km/h*t going south, where t is in hours, meanwhile from the dock the B2 is at a distance of (15km-15km/h*t), t=0, when it is 8pm.

Then we have a right triangle and the distance from boat B1 to boat B2, can be measured as the square root of (15-15*t)^2 +(20*t)^2. We are looking for a minimum, then we have to find the derivative with respect to t. This is 5*(25*t-9)/(sqrt(25*t^2-18*t+9)), this derivative is zero at t=9/25=0,36 h = 21.6 min, now to be sure it is a minimum we apply the second derivative criteria that states that if the second derivative at the given critical point is positive it means here we have a minimum, and by calculating the second derivative we find it is 720/(25 t^2 - 18 t + 9)^(3/2) that is positive at t=9/25, then we have our answer. And besides replacing the value of t we get the distance is 12 km.

3 0
3 years ago
Can you explain that gravity pulls us to the Earth & can you calculate weight from masses on both on Earth and other planets
schepotkina [342]
I don't actually understand what your question is, but I'll dance around the subject
for a while, and hope that you get something out of it.

-- The effect of gravity is:  There's a <em>pair</em> of forces, <em>in both directions</em>, between
every two masses.

-- The strength of the force depends on the <em>product</em> of the masses, so it doesn't matter whether there's a big one and a small one, or whether they're nearly equal. 
It's the product that counts.  Bigger product ==> stronger force, in direct proportion.

-- The strength of the forces also depends on the distance between the objects' centers.  More distance => weaker force.  Actually, (more distance)² ==> weaker force.

-- The forces are <em>equal in both directions</em>.  Your weight on Earth is exactly equal to
the Earth's weight on you.  You can prove that.  Turn your bathroom scale face down
and stand on it.  Now it's measuring the force that attracts the Earth toward you. 
If you put a little mirror down under the numbers, you'll see that it's the same as
the force that attracts you toward the Earth when the scale is right-side-up.

-- When you (or a ball) are up on the roof and step off, the force of gravity that pulls
you (or the ball) toward the Earth causes you (or the ball) to accelerate (fall) toward the Earth. 
Also, the force that attracts the Earth toward you (or the ball) causes the Earth to accelerate (fall) toward you (or the ball).
The forces are equal.  But since the Earth has more mass than you have, you accelerate toward the Earth faster than the Earth accelerates toward you.

--  This works exactly the same for every pair of masses in the universe.  Gravity
is everywhere.  You can't turn it off, and you can't shield anything from it.

-- Sometimes you'll hear about some mysterious way to "defy gravity".  It's not possible to 'defy' gravity, but since we know that it's there, we can work with it.
If we want to move something in the opposite direction from where gravity is pulling it, all we need to do is provide a force in that direction that's stronger than the force of gravity.
I know that sounds complicated, so here are a few examples of how we do it:
-- use arm-muscle force to pick a book UP off the table
-- use leg-muscle force to move your whole body UP the stairs
-- use buoyant force to LIFT a helium balloon or a hot-air balloon 
-- use the force of air resistance to LIFT an airplane.

-- The weight of 1 kilogram of mass on or near the Earth is 9.8 newtons.  (That's
about 2.205 pounds).  The same kilogram of mass has different weights on other planets. Wherever it is, we only know one of the masses ... the kilogram.  In order
to figure out what it weighs there, we need to know the mass of the planet, and
the distance between the kilogram and the center of the planet.

I hope I told you something that you were actually looking for.
7 0
3 years ago
Define Potential Difference ​
GarryVolchara [31]

Answer:

the difference of electrical potential between two points.

Explanation:

3 0
3 years ago
Help pleasessssssssssssss
irinina [24]

Sorry I had the answer but it wont let me type numbers :(.:

5 0
2 years ago
(a) How many fringes appear between the first diffraction-envelope minima to either side of the central maximum in a double-slit
Ainat [17]

Answer:

a

The number of fringe is  z  = 3 fringes

b

The  ratio is I = 0.2545I_o

Explanation:

a

 From the question we are told that

        The wavelength is  \lambda = 600 nm

        The distance between the slit is  d = 0.117mm = 0.117 *10^{-3} m

        The width of the slit is  a = 35.7 \mu m = 35.7 *10^{-6}m

let  z be the number of fringes that appear between the first diffraction-envelope minima to either side of the central maximum in a double-slit pattern is  and this mathematically represented as

             z = \frac{d}{a}

Substituting values

             z = \frac{0.117*10^{-3}}{35.7 *10^{-6}}  

             z  = 3 fringes

b

   From the question  we are told that the order  of the bright fringe is  n = 3

   Generally the intensity of  a pattern  is mathematically represented as

                 I = I_o cos^2 [\frac{\pi d sin \theta}{\lambda} ][\frac{sin (\pi a sin \frac{\theta}{\lambda } )}{\pi a sin \frac{\theta}{\lambda} } ]

Where I_o is the intensity  of the  central fringe

 And  Generally  sin \theta = \frac{n \lambda }{d}

               I = I_o co^2 [ \frac{\pi (\frac{n \lambda}{d} )}{\lambda} ] [\frac{\frac{sin (\pi a (\frac{n \lambda}{d} ))}{\lambda} }{\frac{\pi a (\frac{n \lambda}{d} )}{\lambda} } ]

               I = I_o cos^2 (n \pi)[\frac{\frac{sin(\pi a (\frac{n \lambda}{d} ))}{\lambda} )}{ \frac{ \pi a (\frac{n \lambda }{d} )}{\lambda} } ]

               I = I_o cos^2 (3 \pi) [\frac{sin (\frac{3 \pi }{6} )}{\frac{3 \pi}{6} } ]

                I = I_o (1)(0.2545)

                  I = 0.2545I_o

6 0
3 years ago
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