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Allisa [31]
3 years ago
8

According to the second law of thermodynamics, when energy changes occur, _______ increases. question 2 options:

Physics
1 answer:
Jobisdone [24]3 years ago
7 0
The correct answer is entropy.

In fact, second law of thermodynamics states that the total entropy of an isolated system can never decrease. It can remain constant if the system is in equilibrium or in case of a reversible process: however, perfectly reversible process do not exist in the real world, so the entropy of a system always increases whenever there is a process that involves energy changes.
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A battery is used to charge a parallel-plate capacitor, after which it is disconnected. Then the plates are pulled apart to twic
Igoryamba

Answer: C.

Explanation:

For a parallel-plate capacitor where the distance between the plates is d.

The capacitance is:

C = e*A/d

You can see that the distance is in the denominator, then if we double the distance, the capacitance halves.

Now, the stored energy can be written as:

E = (1/2)*Q^2/C

Now you can see that in this case, the capacitance is in the denominator, then we can rewrite this as:

E = (1/2)*Q^2*d/(e*A)

e is a constant, A is the area of the plates, that is also constant, and Q is the charge, that can not change because the capacitor is disconnected.

Then we can define:

K = (1/2)*Q^2/(e*A)

And now we can write the energy as:

E = K*d

Then the energy is proportional to the distance between the plates, this means that if we double the distance, we also double the energy.

8 0
3 years ago
An engineer is designing a runway for a witch. Several brooms will use the runway and the engineer must design it so that it is
jonny [76]

Answer:

1170 m

Explanation:

Given:

a = 3.30 m/s²

v₀ = 0 m/s

v = 88.0 m/s

x₀ = 0 m

Find:

x

v² = v₀² + 2a(x - x₀)

(88.0 m/s)² = (0 m/s)² + 2 (3.30 m/s²) (x - 0 m)

x = 1173.33 m

Rounded to 3 sig-figs, the runway must be at least 1170 meters long.

6 0
3 years ago
1) Choose the answer choice that BEST completes the following sentence.
Lena [83]

1. one-Half

2. Apogee

3. Any object that revolves around another object

4. Venus's gravitation pull

5 0
3 years ago
Read 2 more answers
What will be the final velocity of a rock if we drop it off of a bridge and it strikes the ground 2.8s later (ignoring air resis
mars1129 [50]
Formula for final velocity: Vf= vi+(a*t)
Vi- initial velocity, a=acceleration, t-time

Vf=vi+(at)
Vf= 0+(9.8m/s*2.8s)
Vf= 27.44 m/s

The acceleration of the Earth when dropping something would be 9.8 m/s

Here is an reference that can help you answer problems like these.
Hope this helps and good luck :)

5 0
2 years ago
A train whistle is heard at 300 Hz as the train approaches town. The train cuts its speed in half as it nears the station, and t
spin [16.1K]

Answer:

The speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.

Explanation:

We can calculate the speed of the train using the Doppler equation:

f = f_{0}\frac{v + v_{o}}{v - v_{s}}        

Where:

f₀: is the emitted frequency

f: is the frequency heard by the observer  

v: is the speed of the sound = 343 m/s

v_{o}: is the speed of the observer = 0 (it is heard in the town)

v_{s}: is the speed of the source =?

The frequency of the train before slowing down is given by:

f_{b} = f_{0}\frac{v}{v - v_{s_{b}}}  (1)                  

Now, the frequency of the train after slowing down is:

f_{a} = f_{0}\frac{v}{v - v_{s_{a}}}   (2)  

Dividing equation (1) by (2) we have:

\frac{f_{b}}{f_{a}} = \frac{f_{0}\frac{v}{v - v_{s_{b}}}}{f_{0}\frac{v}{v - v_{s_{a}}}}

\frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - v_{s_{b}}}   (3)  

Also, we know that the speed of the train when it is slowing down is half the initial speed so:

v_{s_{b}} = 2v_{s_{a}}     (4)

Now, by entering equation (4) into (3) we have:

\frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - 2v_{s_{a}}}  

\frac{300 Hz}{290 Hz} = \frac{343 m/s - v_{s_{a}}}{343 m/s - 2v_{s_{a}}}

By solving the above equation for v_{s_{a}} we can find the speed of the train after slowing down:

v_{s_{a}} = 11.06 m/s

Finally, the speed of the train before slowing down is:

v_{s_{b}} = 11.06 m/s*2 = 22.12 m/s

Therefore, the speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.                        

I hope it helps you!                                                        

7 0
2 years ago
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