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Rainbow [258]
2 years ago
13

As new-forming stars grow by gravitationally attracting more material, they become brighter. If a star becomes too bright, the r

adiation pressure from the star prevents gravity from pulling in more material. What power output PP would a star with a mass 35.135.1 times that of our Sun need to have in order to be unable to pull in a 3.96 μg3.96 μg dust particle of radius rp=0.935 mm,rp=0.935 mm, assuming the particle absorbs all the light incident on its πr2pπrp2 cross‑section? The mass ????⊙M⊙ of the Sun is 1.99×1030 kg,1.99×1030 kg, and the gravitational constant GG equals 6.67×10−11 m3/(kg·s2).
Physics
2 answers:
ratelena [41]2 years ago
4 0

Answer:

Check the explanation

Explanation:

To tackle situations like the one above, the rate of gravity of that star must be equal to the rate of power output but we don’t have radius of that star. Also temp is not mentioned. And emissivity of star is also not mentioned. So the only possible way is like Einstein mass energy relationship E=mc^2=6.5380e40

power =E/Time so this energy is transferred per sec.

SVEN [57.7K]2 years ago
4 0

Answer:

The power output of the star is 2.215x10²⁸W

Explanation:

Given data:

ms = mass of the star = 35.135 times msun

rp = radius = 0.935 mm = 9.35x10⁻⁴m

p = 3.46 μg = 3.46x10⁻⁹g

msun = mass of the Sun = 1.99x10³⁰kg

G = gravitational constant = 6.67x10⁻¹¹m³/kg s²

c = speed of light = 3x10⁸m/s

Question: What power ouput would a star, P = ?

First, calculate the mass of the star:

m_{s} =35.135*1.99x10^{30} =6.992x10^{31}kg

The power output of the star:

P=\frac{Gm_{s}*p*4c }{r^{2} } =\frac{6.67x10^{-11}*6.992x10^{31}*3.46x10^{-9}*4*3x10^{8}}{(9.35x10^{-4})^{2}  } =2.215x10^{28}W

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The angle θ is slowly increased. Write an expression for the angle at which the block begins to move in terms of μs.
Reika [66]

Answer:

tan \theta = \mu_s

Explanation:

An object is at rest along a slope if the net force acting on it is zero. The equation of the forces along the direction parallel to the slope is:

mg sin \theta - \mu_s R =0 (1)

where

mg sin \theta is the component of the weight parallel to the slope, with m being the mass of the object, g the acceleration of gravity, \theta the angle of the slope

\mu_s R is the frictional force, with \mu_s being the coefficient of friction and R the normal reaction of the incline

The equation of the forces along the direction perpendicular to the slope is

R-mg cos \theta = 0

where

R is the normal reaction

mg cos \theta is the component of the weight perpendicular to the slope

Solving for R,

R=mg cos \theta

And substituting into (1)

mg sin \theta - \mu_s mg cos \theta = 0

Re-arranging the equation,

sin \theta = \mu_s cos \theta\\\rightarrow tan \theta = \mu_s

This the condition at which the equilibrium holds: when the tangent of the angle becomes larger than the value of \mu_s, the force of friction is no longer able to balance the component of the weight parallel to the slope, and so the object starts sliding down.

4 0
2 years ago
The suspension cable of a 1,000 kg elevator snaps, sending the elevator moving downward through its shaft. The emergency brakes
tester [92]

Answer:

option (E) 1,000,000 J

Explanation:

Given:

Mass of the suspension cable, m = 1,000 kg

Distance, h = 100 m

Now,

from the work energy theorem

Work done by the gravity = Work done by brake

or

mgh = Work done by brake

where, g is the acceleration due to the gravity = 10 m/s²

or

Work done by brake  = 1000 × 10 × 100

or

Work done by brake = 1,000,000 J

this work done is the release of heat in the brakes

Hence, the correct answer is option (E) 1,000,000 J

4 0
2 years ago
A charge q = 3 × 10-6 C of mass m = 2 × 10-6 kg, and speed v = 5 × 106 m/s enters a uniform magnetic field. The mass experiences
NeX [460]

Answer:

Magnetic field, B = 0.004 mT

Explanation:

It is given that,

Charge, q=3\times 10^{-6}\ C

Mass of charge particle, m=2\times 10^{-6}\ C

Speed, v=5\times 10^{6}\ m/s

Acceleration, a=3\times 10^{4}\ m/s^2

We need to find the minimum magnetic field that would produce such an acceleration. So,

ma=qvB\ sin\theta

For minimum magnetic field,

ma=qvB

B=\dfrac{ma}{qv}

B=\dfrac{2\times 10^{-6}\ C\times 3\times 10^{4}\ m/s^2}{3\times 10^{-6}\ C\times 5\times 10^{6}\ m/s}

B = 0.004 T

or

B = 4 mT

So, the magnetic field produce such an acceleration at 4 mT. Hence, this is the required solution.

4 0
2 years ago
A container of volume 0.6 m^3 contains 5.3 mol of argon gas at 24°C. Assuming argon behaves as an ideal gas, find the total inte
Vitek1552 [10]

Answer:

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Explanation:

let n be the number of moles, R be the gas constant and T be the temperature in Kelvins.

the internal energy of an ideal gas is given by:

Ein = 3/2×n×R×T

     = 3/2×(5.3)×(8.31451)×(24 + 273)

     = 433089.52 J

Therefore, the internal energy of this gas is 433089.52 J.

5 0
2 years ago
If two waves with identical crests and troughs meet, what is happening? The wave is reflecting. Constructive interference is occ
Kazeer [188]
The correct answer would be that destructive interference is happening. In this interference, the crest of a wave meets a trough of another wave resulting to an amplitude that is lower. The opposite is called the constructive interference. Hope this answers the question.
7 0
1 year ago
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