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DerKrebs [107]
3 years ago
15

Photons of light scatter off molecules, and the distance you can see through a gas is proportional to the mean free path of phot

ons through the gas. Photons are not gas molecules, so the mean free path of a photon is not equal to that of a molecule, but its dependence on the number density of the gas and on molecular radius is the same. Suppose you are in a smoggy city and can barely see buildings 500 m away.
(a) How far would you be able to see if all the molecules around you suddenly doubled in volume?
(b) How far would you be able to see if the temperature suddenly rose from 20◦C to a blazing hot 1500◦C with the pressure unchanged?

Physics
1 answer:
Nesterboy [21]3 years ago
6 0

Answer:

a) 315 m

b) 3025.6 m

Explanation:

The picture attached shows the full explanation for the problem.

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Dafna11 [192]
<span>A particular frost-free refrigerator uses about 710kWh of electrical energy per year. You are to express this amount of energy in J, kJ, & Calories. 

1 year (365 days / 1 year)(24 hours / 1 day)(3600s / 1h) = 31,536,000s

710 kWh/yr (1 yr) = 710 kWh
710 x 10^3 Wh = </span>710 x 10^3(J/s)(31,536,000s)<span> = 2.24 x 10^13 J
</span>2.24 x 10^13 J = 2.24 x 10^10 kJ = 5.35 x 10^12 cal
6 0
3 years ago
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One of your summer lunar space camp activities is to launch a 1090 kg rocket from the surface of the Moon. You are a serious spa
Ludmilka [50]

Answer:

ΔP.E = 6.48 x 10⁸ J

Explanation:

First we need to calculate the acceleration due to gravity on the surface of moon:

g = GM/R²

where,

g = acceleration due to gravity on the surface of moon = ?

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

M = Mass of moon = 7.36 x 10²² kg

R = Radius of Moon = 1740 km = 1.74 x 10⁶ m

Therefore,

g = (6.67 x 10⁻¹¹ N.m²/kg²)(7.36 x 10²² kg)/(1.74 x 10⁶ m)²

g = 2.82 m/s²

now the change in gravitational potential energy of rocket is calculated by:

ΔP.E = mgΔh

where,

ΔP.E = Change in Gravitational Potential Energy = ?

m = mass of rocket = 1090 kg

Δh = altitude = 211 km = 2.11 x 10⁵ m

Therefore,

ΔP.E = (1090 kg)(2.82 m/s²)(2.11 x 10⁵ m)

<u>ΔP.E = 6.48 x 10⁸ J</u>

7 0
3 years ago
The key to making a concise mathematical definition of escape velocity is to consider the energy. If an object is launched at it
aleksklad [387]

Answer:

The total Mechanical energy will be zero

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since the object is moving at an escape velocity which is 11.2m/s the object will be assumed to be weightless

Etotal = kinetic energy + potential energy

kinetic energy= 1/2*M*V*V

Potential energy=MGH

Etotal=1/2*0*11.2*11.2+0*0*0

Etotal=0+0

Etotal=0.

3 0
3 years ago
A jet is circling an airport control tower at a distance of 20.6 km. An observer in the tower watches the jet cross in front of
lesya [120]

Answer:

197.76 m

Explanation:

r = Radius of the path = 20.6 km = 20.6\times 10^3\ m

\theta = The angle subtended by moon = 9.6\times 10^{-3}\ rad

Distance traveled is given by

s=r\times\theta

\Rightarrow s=20.6\times 10^3\times 9.6\times 10^{-3}

\Rightarrow s=197.76\ m

The distance traveled by the jet is 197.76 m

8 0
3 years ago
How is force related to math
dybincka [34]

Answer:

Newton's second law of motion describes the relationship between force and acceleration. They are directly proportional. If you increase the force applied to an object, the acceleration of that object increases by the same factor. In short, force equals mass times acceleration.

Explanation:

8 0
3 years ago
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