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3 years ago

Las condiciones iniciales de un gas son 3000 cm3

Physics

1 answer:

slava [35]3 years ago

8 0

Answer:

T'=92.70°C

Explanation:

To find the temperature of the gas you use the equation for ideal gases:

$PV=nRT$

V: volume = 3000cm^3 = 3L

P: pressure = 1250mmHg; 1 mmHg = 0.001315 atm

n: number of moles

R: ideal gas constant = 0.082 atm.L/mol.K

T: temperature = 27°C = 300.15K

For the given values you firs calculate the number n of moles:

$n=\frac{PV}{RT}=\frac{(1520[0.001315atm])(3L)}{(0.082\frac{atm.L}{mol.K})(300.15K)}=0.200moles$

this values of moles must conserve when the other parameter change. Hence, you have V'=2L and P'=3atm. The new temperature is given by:

$T'=\frac{P'V'}{nR}=\frac{(3atm)(2L)}{(0.200\ moles)(0.082\frac{atm.L}{mol.K})}=365.85K=92.70\°C$

hence, T'=92.70°C

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The transfer of energy that occurs when a forceis applied over a distance is _____. Fill in the blank

IceJOKER [234]

Answer:

work

Explanation:

In physics, work is a term that measures the energy expended in a moving object; most commonly force times distance. No work is said to be done if the object does not move. However, during the process, a measure of energy is usually extracted in the process.

4 0

2 years ago

A catapult fires a 0.015kg stone at 70m/s.The spring constant of the catapult is 150N/M. Find the exrtention of the catapult.

adelina 88 [10]

Answer:

0.495m

Explanation:

The kinetic energy of the stone will be equal to the work done by the spring

Mathematically;

W = KE

1/2ke² = 1/2mv²

ke² = mv²

k is the spring constant = 150N/m

e is the extension

m is the mass = 0.015kg

v is the velocity = 70m/s

Substitute

150e² = 1/2 * 0.015 * 70²

150e² = 36.75

e² = 36.75/150

e² = 0.245

e = √0.245

e = 0.495m

Hence the extension of the catapult is 0.495m

7 0

3 years ago

Galileo was a contemporary of

Vitek1552 [10]

Brahe & Kepler

Answer from Quizlet

5 0

2 years ago

The masses of the Earth and Moon are 5.98×1024kg and 7.35×1022kg respectively, and their centers are separated by 3.84×108m.

bija089 [108]

Answer:

C) The Earth-Moon CM follows the orbit around the Sun. Earth and Moon rotate around their CM. The radius of rotation of Moon around the CM is much greater than radius of rotation of Earth around the CM.

Explanation:

At first we assume that both Earth and Moon can be treated as particles, the center of mass of the Earth-Moon system is obtained by using this formula:

$r_{CM} = \frac{r_{E}\cdot m_{E}+r_{M}\cdot m_{M}}{m_{E}+m_{M}}$ (Eq. 1)

Where:

$r_{E}$ - Location of the center of the Earth, measured in kilometers.

$r_{M}$ - Location of the Moon, measured in kilometers.

$r_{CM}$ - Location of the center of mass, measured in kilometers.

$m_{E}$ - Mass of the Earth, measured in kilograms.

$m_{M}$ - Mass of the Moon, measured in kilograms.

If we know that $r_{E} = 0\,km$ , $r_{M} = 3.84\times 10^{8}\,m$ , $m_{E} = 5.98\times 10^{24}\,kg$ and $m_{M} = 7.35\times 10^{22}\,kg$ , the location of the center of mass respect to the Earth is:

$r_{CM} = \frac{(0\,km)\cdot (5.98\times 10^{22}\,kg)+(3.84\times 10^{8}\,m)\cdot (7.35\times 10^{22}\,kg)}{5.98\times 10^{24}\,kg+7.35\times 10^{22}\,kg}$

$r_{CM} = 4.662\times 10^{6}\,m$

The Earth has a radius of $6.371\times 10^{6}$ meters, we notice that center of mass in located inside the Earth and the radius of rotation of the Earth around the center of mass is much greater than the radius of rotation of the Moon around the center of mass. That center of mass follows an orbit around the sun.

In consequence, correct answer is C.

4 0

3 years ago

Explain what happens inside the radiation zone that causes the photon to take so long to exit the sun

DaniilM [7]

-A photon travels, on average, a particular distance, d, before being briefly absorbed and released by an atom, which scatters it in a new random direction.

-Given d and the speed of light, c, you can figure out the average time step and space step size (how often the photon “steps” and how far it “steps” each time).

-The size of the Sun is figured in terms of step size. Some surprisingly tricky math happens, involving “Brownian motion” and probabilities. Finally,

-The average time it would take to get to the surface of the Sun is found.

6 0

3 years ago