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larisa [96]
3 years ago
11

What state of matter takes the shape and volume of its container, like the steam coming from a pot A) gas B) liquid Eliminate C)

plasma D) solid
Physics
2 answers:
Gnoma [55]3 years ago
6 0

Ummm gas takes the shape and volume of its container, but I don't know about plasma. Solids definitely don't, and liquids don't change their volumes. So I would say gas. :)

BARSIC [14]3 years ago
4 0

It said “like the steam coming from a pot” which probably means it’s A) gas
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Answer:

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R=0.08206 L.atm /mol/k

760mmHg = 1 atm therefore

600mmHg = X atm

760 X = 600mmHg

X = 600/760 = 0.789 atm

P = 0.789 atm

V= ?

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0.789 V = 2.5 × 0.08206 × 298

V= 2.5 × 0.08206 ×298 / 0.789

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Even though you are getting closer to the sun as you ascend into the toposphere, why does the temperature drop?
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igor_vitrenko [27]

Answer: (a) α = \frac{3I}{2.\pi.R^{3}}

(b) For r≤R: B(r) = μ_0.(\frac{I.r^{2}}{2.\pi.R^{3}})

For r≥R: B(r) = μ_0.(\frac{I}{2.\pi.r})

Explanation:

(a) The current I enclosed in a straight wire with current density not constant is calculated by:

I_{c} = \int {J} \, dA

where:

dA is the cross section.

In this case, a circular cross section of radius R, so it translates as:

I_{c} = \int\limits^R_0 {\alpha.r.2.\pi.r } \, dr

I_{c} = 2.\pi.\alpha \int\limits^R_0 {r^{2}} \, dr

I_{c} = 2.\pi.\alpha.\frac{r^{3}}{3}

\alpha = \frac{3I}{2.\pi.R^{3}}

For these circunstances, α = \frac{3I}{2.\pi.R^{3}}

(b) <u>Ampere's</u> <u>Law</u> to calculate magnetic field B is given by:

\int\ {B} \, dl = μ_0.I_{c}

(i) First, first find I_{c} for r ≤ R:

I_{c} = \int\limits^r_0 {\alpha.r.2\pi.r} \, dr

I_{c} = 2.\pi.\frac{3I}{2.\pi.R^{3}} \int\limits^r_0 {r^{2}} \, dr

I_{c} = \frac{I}{R^{3}}\int\limits^r_0 {r^{2}} \, dr

I_{c} = \frac{3I}{R^{3}}\frac{r^{3}}{3}

I_{c} = \frac{I.r^{3}}{R^{3}}

Calculating B(r), using Ampere's Law:

\int\ {B} \, dl = μ_0.I_{c}

B.2.\pi.r = (\frac{Ir^{3}}{R^{3}} ).μ_0

B(r) = (\frac{Ir^{3}}{R^{3}2.\pi.r}).μ_0

B(r) = (\frac{Ir^{2}}{2.\pi.R^{3}} ).μ_0

For r ≤ R, magnetic field is B(r) = (\frac{Ir^{2}}{2.\pi.R^{3}} ).μ_0

(ii) For r ≥ R:

I_{c} = \int\limits^R_0 {\alpha.2,\pi.r.r} \, dr

So, as calculated before:

I_{c} = \frac{3I}{R^{3}}\frac{R^{3}}{3}

I_{c} = I

Using Ampere:

B.2.π.r = μ_0.I

B(r) = (\frac{I}{2.\pi.r} ).μ_0

For r ≥ R, magnetic field is; B(r) = (\frac{I}{2.\pi.r} ).μ_0.

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3 years ago
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