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melisa1 [442]
2 years ago
14

Suppose an isolated box of volume 2V is divided into two equal compartments. An ideal gas occupies half of the container and the

other half is empty. When the partition separating the two halves of the box is removed and the system reaches equilibrium again, how does the new internal energy of the gas compare to the internal energy of original system? a. The internal energy stays the same b. There is not enough information to determine the answer c. The internal energy decreases d. The internal energy increases
Physics
1 answer:
SpyIntel [72]2 years ago
7 0

Answer:

A. the internal energy stays the same

Explanation:

From the first law of thermodynamics, "energy can neither be created nor destroyed but can be transformed from one form to another.

Based on this first law of thermodynamic, the new internal energy of the gas is the same as the internal energy of the original system.

Therefore, when the partition separating the two halves of the box is removed and the system reaches equilibrium again, the internal energy stays the same.

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A thin, horizontal, 18-cm-diameter copper plate is charged to -3.8 nC. Assume that the electrons are uniformly distributed on th
son4ous [18]

Answer:

Part a)

E = 8436.7 N/C

Part b)

E = 8436.7 N/C

Explanation:

Part a)

Electric field due to large sheet is given as

E = \frac{\sigma}{2\epsilon_0}

\sigma = \frac{Q}{A}

Q = -3.8 nC

A = \pi(0.09)^2

A = 0.025 m^2

\sigma = \frac{-3.8\times 10^{-9}}{0.025}

\sigma = -1.5 \times 10^{-7} C/m^2

now the electric field is given as

E = \frac{-1.5 \times 10^{-7}}{2(8.85 \times 10^{-12})}

E = 8436.7 N/C

Part b)

Now since the electric field is required at same distance on other side

so the field will remain same on other side of the plate

E = 8436.7 N/C

5 0
2 years ago
A turntable rotates counterclockwise at 76 rpm . A speck of dust on the turntable is at 0.47 rad at t=0. What is the angle of th
icang [17]

To solve this exercise it is necessary to apply the kinematic equations of angular motion.

By definition we know that the displacement when there is constant angular velocity is

\theta= \theta_0 +\omega t

From our given data we know that,

\omega = 76\frac{rev}{min}

\omega = 76\frac{rev}{min}(\frac{2\pi rad}{1rev})(\frac{1 min}{60s})

\omega = 7.958rad/s

Moreover we know that

\theta_0 = 0.47 rad

Therefore for time t=8.1s we have,

\theta= \theta_0+ \omega t

\theta= 0.47+(7.958)(8.1)

\theta = 64.9298rad

That number in revolution is:

\theta = 64.9298rad(\frac{1rev}{2\pi})

\theta = 15.108 Revolutions

Here, we see that there are 15 complete revolutions

And 0.108 revolutions i not complete, so the tunable rotation is

\theta_{net} = 0.108*2\pi=0.216\pi

Therefore the angle of the speck at a time 8.1s is 0.216\pi

4 0
2 years ago
You need to determine the density of an unknown liquid and decide to perform an experiment. You notice that a wooden block float
Allushta [10]

Answer:

pu = 1260.9kg/m^3

the density of the unknown liquid is 1260.9kg/m^3

Explanation:

The density of a liquid is inversely proportional to the volume (height) of object submerged in it.

High density liquid possess higher buoyant force preventing objects from submerging.

p ∝ 1/V ∝ 1/h

since V = Ah

pu/pw = hw/hu

pu = pwhw/hu

Where;

p = density

h = height submerged

pu and pw is the density of unknown liquid and water respectively

hu and hw is the height of object submerged in unknown liquid and water respectively

pw = 1000kg/m^3

hu = 4.6cm = 0.046m

hw = 5.8cm = 0.058m

Substituting the given values;

pu = 1000×0.058/0.046

pu = 1260.9kg/m^3

the density of the unknown liquid is 1260.9kg/m^3

5 0
2 years ago
A 5 kg object near Earth's surface is released from rest such that it falls a distance of 10 m. After the object falls 10 m, it
makkiz [27]

Answer:D

Explanation:

Given

mass of object m=5 kg

Distance traveled h=10 m

velocity acquired v=12 m/s

conserving Energy at the moment when object start falling and when it gains 12 m/s velocity

Initial Energy=mgh=5\times 9.8\times 10=490 J

Final Energy=\frac{1}{2}mv^2+W_{f}

=\frac{1}{2}\cdot 5\cdot 12^2+W_{f}

where W_{f} is friction work if any

490=360+W_{f}

W_{f}=130 J

Since Friction is Present therefore it is a case of Open system and net external Force is zero

An open system is a system where exchange of energy and mass is allowed and Friction is acting on the object shows that system is Open .

4 0
2 years ago
Car B is following Car A and has a greater speed than Car A. the two cars are moving in a straight line and in the same directio
ElenaW [278]

Answer:

Collision force will be same in both the cases.

Explanation:

A perfectly inelastic collision is said to take place  when a system loses the amount of its Kinetic Energy at its maximum. In a perfectly inelastic collision,  the colliding particles stick to each other. In such a collision, kinetic energy is lost by combining the two bodies with each other.

In situation 1:

Speed of Car A, v_{A} = 10 mph

Speed of Car B, v_{B} = 20 mph

Relative speed of car A and car B, v = v_{b} - v_{a} = 10 m/s

Now, in the situation 2:

Speed of car A, v_{A} = 30 mph

Speed of car B, v_{B} = 40 mph

Relative speed of car A and car B, v = v_{b} - v_{a} = 10 m/s

Therefore, Car A and Car B both have the same relative speed, v = 10 m/s

7 0
2 years ago
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