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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 years ago
What is the velocity of Car B after collision?
zubka84 [21]

Answer:

30m/s

Explanation:

m1v1=m2v2

12 * 10=4* v2

120/4 =v2

V2=30m/s

7 0
3 years ago
(1 point) A projectile is fired from ground level with an initial speed of 600 m/sec and an angle of elevation of 30 degrees. Us
konstantin123 [22]

Answer:

(a) The range of the projectile is 31,813.18 m

(b) The maximum height of the projectile is 4,591.84 m

(c) The speed with which the projectile hits the ground is 670.82 m/s.

Explanation:

Given;

initial speed of the projectile, u = 600 m/s

angle of projection, θ = 30⁰

acceleration due to gravity, g = 9.8 m/s²

(a) The range of the projectile in meters;

R = \frac{u^2sin \ 2\theta}{g} \\\\R = \frac{600^2 sin(2\times 30^0)}{9.8} \\\\R = \frac{600^2 sin (60^0)}{9.8} \\\\R = 31,813.18 \ m

(b) The maximum height of the projectile in meters;

H = \frac{u^2 sin^2\theta}{2g} \\\\H = \frac{600^2 (sin \ 30)^2}{2\times 9.8} \\\\H = \frac{600^2 (0.5)^2}{19.6} \\\\H = 4,591.84 \ m

(c) The speed with which the projectile hits the ground is;

v^2 = u^2 + 2gh\\\\v^2 = 600^2 + (2 \times 9.8)(4,591.84)\\\\v^2 = 360,000 + 90,000.064\\\\v = \sqrt{450,000.064} \\\\v = 670.82 \ m/s

5 0
3 years ago
An arrow movirg 48.3 m/s has 5.22<br> kg•m/s of momentum. What is its<br> mass?
spin [16.1K]

Answer:

0.11 kg

Explanation:

Ft = MV

Ft = momentum 5.22kg m/s

M = mass

V = velocity 48.3m/s

Therefore

5.22 = M x 48.3

Divide both sides by 48.3

5.22/48.3 = M x 48.3/48.3

0.11 = M

M = 0.11kg

6 0
3 years ago
A battery-operated car utilizes a 120.0 V battery with negligible internal resistance. Find the charge, in coulombs, the batteri
fgiga [73]

Answer:

4.29×10⁵ C

Explanation:

From the question,

The energy stored in the battery = Kinetic energy of the car+ Energy needed to make the car climbed the hill+Energy required to exert a force.

E = 1/2mv²+mgh+Fd.................... Equation 1

Where E = Energy stored in the battery, m = mass of the car, v = velocity of the car, h = height of the hill, F = force exerted on the car, d = distance traveled by the car.

But,

d = vt.................... Equation 2

Where v = velocity, t = time.

Substitute equation 2 into equation 1

E = 1/2mv²+mgh+F(vt)................... Equation 3

Given: m = 770 kg, v = 26 m/s, h = 2.15×10² m = 215 m, F = 5.3×10² N = 530 N, t = 1 hour = 3600 s, g = 9.8 m/s²

Substitute into equation 1

E = 1/2(770)(26²)+(770)(9.8)(215)+(530)(26)(3600)

E = 260260+1622390+49608000

E = 51490650 J

Using,

E = qV................. Equation 4

Where q = charge of the battery, V = Voltage.

make q the subject of the equation

q = E/V............... Equation 5

Given: E = 51490650 J, V = 120 V

Substitute into equation 5

q = 51490650/120

q = 429088.75 C

q = 4.29×10⁵ C

7 0
4 years ago
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