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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 sound technician is testing the sound acoustics in a theatre for an upcoming music concert. As he moves towards the speakers,
Harlamova29_29 [7]

Answer: Increase in wave frequency

Explanation:

When we talk about acoustics we are dealing with sound waves, and one of their main components along with the velocity and wavelength is the <u>frequency.</u>

In this sense, the frequency of any wave refers to how fast (or slow) a wave oscillates. For example, in the especific case of sound waves when the oscillation is faster, the frequency is higher and the pitch gets higher as well.

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What can happen to an electron when sunlight hits it? select all that apply. select all that apply. it can drop down to a lower
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An archer fires an arrow, which produces a muffled "thwok" as it hits a target. If the archer hears the "thwok" exactly 1 s afte
aniked [119]

Answer:

35,79 meters

Explanation:

So, we got an archer, and we got a target. Lets call the distance between this two d.

Now, the archer fires the arrow, that, in a time t_{arrow} travels the distance d with a speed v_{arrow} of 40 m/s and hits the target. We can see that the equation will be:

v_{arrow} * t_{arrow} = d\\ \\40 \frac{m}{s} * t_{arrow} = d

Immediately after this, the arrow produces a muffled sound, which will travel the distance d at  340 m/s in a time t_{sound}. Obtaining :

v_{sound} * t_{sound} = d\\ \\340 \frac{m}{s} * t_{sound} = d.

Finally, the sound reaches the archer, exactly 1 second after he fired the bow, so:

t_{arrow} + t _{sound} = 1 s.

This equation allows us to write:

t _{sound} = 1 s - t_{arrow}.

Plugging this  relationship in the distance equation for the sound:

340 \frac{m}{s} * t_{sound} = d \\ \\ 340 \frac{m}{s} * (1 s- t_{arrow}) = d.

Now, we can replace d from the first equation, and obtain:

40 \frac{m}{s} * t_{arrow} = d \\ 40 \frac{m}{s} * t_{arrow} = 340 \frac{m}{s} * (1 s- t_{arrow}).

Now, we can just work a little bit:

40 \frac{m}{s} * t_{arrow} = 340 \frac{m}{s} * 1 s - 340 \frac{m}{s} * t_{arrow} \\ \\ 40 \frac{m}{s} * t_{arrow} + 340 \frac{m}{s} * t_{arrow} = 340 m \\ \\ 380 \frac{m}{s} * t_{arrow} = 340 m \\ \\ t_{arrow} = \frac{340 m}{380 \frac{m}{s}} \\ \\ t_{arrow} = 0.8947 s.

Now, we can just plug this value into the first equation:

40 \frac{m}{s} * t_{arrow} = d

40 \frac{m}{s} * 340/380 s = 35,79 s = d

6 0
2 years ago
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