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wel
4 years ago
6

Approximately 1.000 g each of four gasses H2, Ne, Ar, and Kr are placed in a sealed container all under1.5 atm of pressure. Assu

ming ideal behavior, determine the partial pressure of the H2 and Ne
Physics
1 answer:
Vera_Pavlovna [14]4 years ago
7 0

Answer:

The partial pressure of H2 is 0.375 atm

The partial pressure of Ne is also 0.375 atm

Explanation:

Mass of H2 = 1 g

Mass of Ne = 1 g

Mass of Ar = 1 g

Mass of Kr = 1 g

Total mass of gas mixture = 1 + 1 + 1 + 1 = 4 g

Pressure of sealed container = 1.5 atm

Partial pressure of H2 = (mass of H2/total mass of gas mixture) × pressure of sealed container = 1/4 × 1.5 = 0.375 atm

Partial pressure of Ne = (mass of Ne/total mass of gas mixture) × pressure of sealed container = 1/4 × 1.5 = 0.375 atm

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A) +5 J

B) +1 J

Explanation:

A)

The internal forces (interaction forces) acting on a system do not change the mechanical energy (sum of potential and kinetic energy) of the system.

However, these forces are responsible for converting the energy from one form into another; the work done by these forces is equal to the amount of energy converted from one form into the other.

In this problem, we have:

\Delta U=-5 J is the loss in potential energy of the system

\Delta K=+6 J is the gain in kinetic energy of the system

By looking at these numbers, this means that the internal forces have converted 5 J of energy from potential energy into kinetic energy (while the additional +1 J missing is due to external forces, as explained in part B).

Therefore, the work done by internal forces is

W = +5 J

B)

First of all, we calculate the change in mechanical energy of the system.

The mechanical energy of a system is the sum of its kinetic energy (K) and its potential energy (U):

E=K+U

So, the change in mechanical energy is equal to the sum of the changes of kinetic energy and the changes of potential energy:

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In this problem:

\Delta K=+6 J

\Delta U=-5 J

So, the change in mechanical energy is:

\Delta E=+6+(-5)=+1 J

According to the work-energy theorem, the work done by external forces on a system is equal to the change in mechanical energy of the system: therefore in this case, the work done by external forces is

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4 years ago
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KonstantinChe [14]

Answer:

c. metallic, sub metallic, or nonmetallic

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3 years ago
A piece of Nichrome wire has a radius of 6.5 104 m. It is used in a laboratory to make a heater that uses 4.00 102 W of power wh
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Answer:

L=4.8*10^{17}m

Explanation:

Given data

Power P=4.00×10²W

Radius r=6.5×10⁴m

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Length of wire L

Solution

We know that resistance of wire can be obtained from

P=\frac{V_{2}}{R}\\ R=\frac{V_{2}}{P}

We also know that R=pL/A solving the length noting that A=πr²

and using p=100×10⁻⁸Ω.m we find that

So

L=\frac{RA}{p}\\ L=\frac{\frac{(V^{2})}{P}(\pi r^{2}) }{p} \\L=\frac{V^{2}(\pi r^{2})}{pP}\\ L=\frac{(120V)^{2}\pi (6.5*10^{4} m)^{2}  }{100*10^{-8}(4.00*10^{2} W) }\\ L=4.8*10^{17}m

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andrew-mc [135]

Answer:

Explanation:

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6 0
3 years ago
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