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2 years ago

Which of the following situations would violate the second law of thermodynamics?

Physics

2 answers:

Fiesta28 [93]2 years ago

6 0

Answer:

A heat engine feeling cold after running for an hour -appex

Explanation:

Musya8 [376]2 years ago

4 0

Heat flows irreversibly from hot to cold

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A brass lid screws tightly onto a glass jar at 20 degrees C. To help open the jar, it can be placed into a bath of hot water. Af

omeli [17]

Answer:

0.0016 cm

Explanation:

$\alpha_b$ = Thermal coefficient of expansion of brass = $19\times 10^{-6}\ /^{\circ}C$

$\alpha_g$ = Thermal coefficient of expansion of glass = $9\times 10^{-6}\ /^{\circ}C$

$\Delta T$ = Change in temperature = $(60-20)^{\circ}C$

$R_0$ = Initial radius = 4 cm

Change in radius of material is given by

$R=R_0(1+\alpha\Delta T)$

Difference in radii of the lid and jar

$\Delta R=R_b-R_g\\\Rightarrow \Delta R=R_0(1+\alpha_b\Delta T)-R_0(1+\alpha_g\Delta T)\\\Rightarrow \Delta R=R_0(\alpha_b-\alpha_g)\Delta T\\\Rightarrow \Delta R=4\times (19\times 10^{-6}-9\times 10^{-6})\times (60-20)\\\Rightarrow \Delta R=0.0016\ cm$

The size of the gap is 0.0016 cm or 0.000016 m

8 0

2 years ago

A solid cylindrical bar conducts heat at a rate of 25 W from a hot to a cold reservoir under steady state conditions. If both th

expeople1 [14]

Answer:

Using the new cylinder the heat rate between the reservoirs would be 50 W

Explanation:

Conduction could be described by the Law of Fourierin the form: $Q=kA\frac{T_1-T_2}{L}$ where $Q$ is the rate of heat transferred by conduction, $k$ is the thermal conductivity of the material, $T_1$ and $T_2$ are the temperatures of each heat deposit, $A$ is the cross area to the flow of heat, and ${L}$ is the distance that the flow of heat has to go.
For the original cylinder the Fourier's law would be: $kA_1\frac{T_1-T_2}{L_1}=25W$ , and if $A_1=\frac{\pi D_{1}^{2}}{4}$ , then the expression would be: $k\frac{\pi D_1^{2}}{4} \frac{T_1-T_2}{L_1}=25W$ where $D_1$ is the diameter of the original cylinder, and ${L_1}$ is the length of the original cylinder.
For the new cylinder, in the same fashion that for the first, Fourier's Law would be: $Q_2=k\frac{\pi D_2^2}{4}\frac{T_1-T_2}{L_2}$ ,where $Q_2$ is the heat rate in the second case, $D_2$ and ${L_2$ are the new diameter and length.
But, $D_2=2D_1$ and $L_2=2L_1$ , substituting in the expression for $Q_2$ : $Q_2=k\frac{\pi (2D_1)^2}{4}\frac{T_1-T_2}{2L_1}$ .
Rearranging: $Q_2=\frac{2^2}{2}(k\frac{\pi D_1^2}{4}\frac{T_1-T_2}{L_1})$ .
In the last declaration of $Q_2$ , it could be noted that the expressión inside the parenthesis is actually $Q_1$ , then: $Q_2=\frac{2^2}{2}(25W)=50W$ .
<u>It should be noted, that the temperatures in the hot and cold reservoirs never change.</u>

7 0

2 years ago

Where is the steering nozzle located on a pwc?

Dvinal [7]

At the rear.

PWC stands for personal watercraft, and it is a small powerboat. The main components of a PWC are the hull (body of the boat), deck (surface where people walk/stand), throttle (controls speed), steering nozzle and water intake.

3 0

2 years ago

Read 2 more answers

You start with spring that's already been stretched an unknown amount from equilibrium. After stretching it an additional 2.0 cm

maxonik [38]

Answer: 35*10^3 N/m

Explanation: In order to explain this problem we know that the potential energy for spring is given by:

Up=1/2*k*x^2 where k is the spring constant and x is the streching or compresion position from the equilibrium point for the spring.

We also know that with additional streching of 2 cm of teh spring, the potential energy is 18J. Then it applied another additional streching of 2 cm and the energy is 25J.

Then the difference of energy for both cases is 7 J so:

ΔUp= 1/2*k* (0.02)^2 then

k=2*7/(0.02)^2=35000 N/m

7 0

2 years ago

An astronaut is in equilibrium when he is positioned 140 km from the center of asteroid C and 581 km from the center of asteroid

tekilochka [14]

Answer:B

Explanation:

Given

Distance of astronaut From asteroid x is $r_x=140 km$