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3 years ago

HELP PLZ! does the distance between particles of a liquid increase as the liquid in a thermometer increases?

Physics

2 answers:

yuradex [85]3 years ago

6 0

Explanation:

The distance between particles increases, and the alcohol expands. As the alcohol expands, its volume increases. Alcohol pushes up the stem of the thermometer. The greater the kinetic energy of the alcohol particles, the more the alcohol expands.

Kazeer [188]3 years ago

4 0

Answer:

The distance between particles increases, and the alcohol expands.

Explanation:

welp just realized thats what the other person said

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True or False: Radio waves are the longest and have the least energy of any wave.

Alona [7]

Answer:

Explanation:

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3 years ago

Read 2 more answers

Two planets P1 and P2 orbit around a star S in circular orbits with speeds v1 = 40.2 km/s, and v2 = 56.0 km/s respectively. If t

Readme [11.4K]

Answer: $3.66(10)^{33}kg$

Explanation:

We are told both planets describe a circular orbit around the star S. So, let's approach this problem begining with the angular velocity $\omega$ of the planet P1 with a period $T=750years=2.36(10)^{10}s$ :

$\omega=\frac{2\pi}{T}=\frac{V_{1}}{R}$ (1)

Where:

$V_{1}=40.2km/s=40200m/s$ is the velocity of planet P1

$R$ is the radius of the orbit of planet P1

Finding $R$ :

$R=\frac{V_{1}}{2\pi}T$ (2)

$R=\frac{40200m/s}{2\pi}2.36(10)^{10}s$ (3)

$R=1.5132(10)^{14}m$ (4)

On the other hand, we know the gravitational force $F$ between the star S with mass $M$ and the planet P1 with mass $m$ is:

$F=G\frac{Mm}{R^{2}}$ (5)

Where $G$ is the Gravitational Constant and its value is $6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$

In addition, the centripetal force $F_{c}$ exerted on the planet is:

$F_{c}=\frac{m{V_{1}}^{2}}{R^{2}}$ (6)

Assuming this system is in equilibrium:

$F=F_{c}$ (7)

Substituting (5) and (6) in (7):

$G\frac{Mm}{R^{2}}=\frac{m{V_{1}}^{2}}{R^{2}}$ (8)

Finding $M$ :

$M=\frac{V^{2}R}{G}$ (9)

$M=\frac{(40200m/s)^{2}(1.5132(10)^{14}m)}{6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}}$ (10)

Finally:

$M=3.66(10)^{33}kg$ (11) This is the mass of the star S

4 0

4 years ago

Suppose Gabor, a scuba diver, is at a depth of 15m. Assume that: The air pressure in his air tract is the same as the net water

s2008m [1.1K]

Complete Question

The complete question is shown on the first and second uploaded image

Answer:

Now the ratio of the gases in Gabor's lungs at the depth of 15m to that at

the surface is $\frac{(n/V)_{15\ m}}{(n/V)_{surface}} = 2.5$

The number of moles of gas that must be released is $n= 0.3538\ mols$

Explanation:

We are told from the question that the pressure at the surface is 1 atm and for each depth of 10m below the surface the pressure increase by 1 atm

This means that the pressure at the depth of the surface would be

$P_d = [\frac{15m}{10m} ] (1 atm) + 1 atm$

$= 2.5 atm$

The ideal gas equation is mathematically represented as

$PV = nRT$

Where P is pressure at the surface

V is the volume

R is the gas constant = 8.314 J/mol. K

making n the subject we have

$n = \frac{PV}{RT}$

Considering at the surface of the water the number of moles at the surface would be

$n_s = \frac{P_sV}{RT}$

Substituting $1 atm = 101325 N/m^2$ for $P_s$ , $6L = 6*10^{-3}m^3$ for volume , 8.314 J/mol. K for R , (37° +273) K for T into the equation

$n_s = \frac{(1atm)(6*10^{-3} m^3)}{(8.314J/mol \cdot K)(37 +273)K}$

$= 0.2359 mol$

To obtain the number of moles at the depth of the water we use

$n_d = \frac{P_d V}{RT}$

Where $P_d \ and \ n_d \$ are pressure and no of moles at the depth of the water

Substituting values we have

$n_d = \frac{(2.5)(101325 N/m^2)(6*10^{-3}m^3)}{(8.314 J/mol \cdot K)(37 + 273)K}$

$= 0.5897 mol$

Now to obtain the number of moles released we have

$n = n_d - n_s$

$= 0.5897mol - 0.2359mol$

$=0.3538 \ mol$

The molar concentration at the surface of water is

$[\frac{n}{V} ]_{surface} = \frac{0.2359mol}{6*10^-3m^3}$

$=39.31mol/m^3$

The molar concentration at the depth of water is

$[\frac{n}{V} ]_{15m} = \frac{0.5897}{6*10^{-3}}$

$= 98.28 mol/m^3$

Now the ratio of the gases in Gabor's lungs at the depth of 15m to that at the surface is

$\frac{(n/V)_{15\ m}}{(n/V)_{surface}} = \frac{98.28}{39.31} =2.5$

6 0

3 years ago

What are the importance of physics

Alexus [3.1K]

Answer:

Physics contributes to the technological infrastructure and provides trained personnel needed to take advantage of scientific advances and discoveries. Physics is an important element in the education of chemists, engineers and computer scientists, as well as practitioners of the other physical and biomedical sciences.

Explanation:

4 0

3 years ago

Two acrobats flying through the air grab and hold onto each other in midair as part of a circus act.One acrobat has a mass of 60

earnstyle [38]

Answer:

1.36m/s

Explanation:

We are given that

Mass of one acrobat, $m_1=60 kg$