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weqwewe [10]
3 years ago
14

Equal masses of he and ne are placed in a sealed container. What is the partial pressure of he if the total pressure in the cont

ainer is 6 atm?
Physics
1 answer:
EastWind [94]3 years ago
6 0

Answer:

6 atm.

Explanation:

Let the mass of both be m

Then moles of He = m/ 4

Moles of Ne = m/ 20

mole fraction of He = Moles of He/ Total moles = m/4/ (m/4 + m/20) = 0.25 m/0.3m = 0.83

Pressure of He = Mole fraction×total pressure = 0.83 × 6 atm = 5 atm

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Meg walks with a velocity of 0.9 m/s west. She does so while riding on a train that is traveling with a velocity of 2.7 m/s east
Shkiper50 [21]
<span>Velocities are vectors so we can add them!

Let's let +x be East and -x be West.

 -0.9 + 2.7 = 1.8
 
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The speed of light in air is 3 x 108 m/s. The speed of light in ice is 2.29 x 108 m/s. What is the refractive index from air to
Studentka2010 [4]

Answer:

η = 1.31

Explanation:

The formula for the refractive index of from air to some other medium is given by the following formula:

\eta = \frac{c}{v}\\

where,

η = refractive index = ?

c = speed of light in air = 3 x 10⁸ m/s

v = speed of light in ice = 2.29 x 10⁸ m/s

Therefore, using these values in the equation we get:

\eta = \frac{3\ x\ 10^8\ m/s}{2.29\ x\ 10^8\ m/s} \\

<u>η = 1.31</u>

4 0
2 years ago
Which statement about cellulose is true?
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In 2005 astronomers announced the discovery of a large black hole in the galaxy Markarian 766 having clumps of matter orbiting a
IRISSAK [1]

A. 4.64\cdot 10^{11}m

The orbital speed of the clumps of matter around the black hole is equal to the ratio between the circumference of the orbit and the period of revolution:

v=\frac{2\pi r}{T}

where we have:

v=30,000 km/s = 3\cdot 10^7 m/s is the orbital speed

r is the orbital radius

T=27 h \cdot 3600 =97,200 s is the orbital period

Solving for r, we find the distance of the clumps of matter from the centre of the black hole:

r=\frac{vT}{2\pi}=\frac{(3\cdot 10^7 m/s)(97200 s)}{2\pi}=4.64\cdot 10^{11}m

B. 6.26\cdot 10^{36}kg, 3.13\cdot 10^6 M_s

The gravitational force between the black hole and the clumps of matter provides the centripetal force that keeps the matter in circular motion:

m\frac{v^2}{r}=\frac{GMm}{r^2}

where

m is the mass of the clumps of matter

G is the gravitational constant

M is the mass of the black hole

Solving the formula for M, we find the mass of the black hole:

M=\frac{v^2 r}{G}=\frac{(3\cdot 10^7 m/s)^2(4.64\cdot 10^{11} m)}{6.67\cdot 10^{-11}}=6.26\cdot 10^{36}kg

and considering the value of the solar mass

M_s = 2\cdot 10^{30}kg

the mass of the black hole as a multiple of our sun's mass is

M=\frac{6.26\cdot 10^{36} kg}{2\cdot 10^{30} kg}=3.13\cdot 10^6 M_s

C. 9.28\cdot 10^9 m

The radius of the event horizon is equal to the Schwarzschild radius of the black hole, which is given by

R=\frac{2MG}{c^2}

where M is the mass of the black hole and c is the speed of light.

Substituting numbers into the formula, we find

R=\frac{6.26\cdot 10^{36} kg)(6.67\cdot 10^{-11})}{(3\cdot 10^8 m/s)^2}=9.28\cdot 10^9 m

8 0
3 years ago
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