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ivanzaharov [21]

2 years ago

11

Most stars are main-sequence stars, a group of stars for which size, mass, surface temperature, and radiated power are closely r

elated. The sun, for instance, is a yellow main sequence star with a surface temperature of 5800 K. For a main-sequence star whose mass M is more than twice that of the sun, the total radiated power, relative to the sun, is approximately P/Psun = 1.5(M/Msun)3.5. The star Regulus A is a bluish main-sequence star with mass 3.8Msun and radius 3.1Rsun. What is the surface temperature of Regulus A?

1 answer:

quester [9]2 years ago

4 0

Answer:

the surface temperature of Regulus A is 11724.13 K

Explanation:

Given the data in the question;

Sun's surface temperature T = 5800 K

total radiated power, relative to the sun is; P/P $_{sun$ = $($ 1.5(M/M $_{sun$ $)^{3.5$

The star Regulus A is a bluish main-sequence star with mass 3.8M $_{sun$ and radius 3.1R $_{sun$ .

First, we determine the value power emitted by the sun or sun as follows;

P = eσAT⁴

where P is the power, e is surface emissivity, σ is Stefan Boltzmann, A is area and T is temperature.

so, lets assume emissivity of star and sun is same;

let p be power related to star and p $_{sun$ be power related sun.

Ratio of power radiated by star and power radiated by sun;

P/P $_{sun$ = eσAT⁴ / eσA $_{sun$ T $_{sun$ ⁴

we know that AREA A = πR²

we input the formula for area

P/P $_{sun$ = eσ(πR²)T⁴ / eσ(π(R $_{sun$ )²)T $_{sun$ ⁴

such that we now have;

P/P $_{sun$ = R²T⁴ / R $_{sun$ ²T $_{sun$ ⁴

given that P/P $_{sun$ = $($ 1.5(M/M $_{sun$ $)^{3.5$ , we substitute

$($ 1.5(M/M $_{sun$ $)^{3.5$ = R²T⁴ / R $_{sun$ ²T $_{sun$ ⁴

we find temperature of the star T

T = 5800 × $[$ 1.5 $($ M/M $_{sun$ $)^{3.5$ (R $_{sun$ ²/R²) $]^{1/4$

Given that; mass M is 3.8M $_{sun$ and radius R is 3.1R $_{sun$ .

we substitute

T = 5800 × $[$ 1.5 $($ 3.8M $_{sun$ /M $_{sun$ $)^{3.5$ (R $_{sun$ ²/( 3.1R $_{sun$ )²) $]^{1/4$

T = 5800 × $[$ 1.5 $($ 3.8 $)^{3.5$ ( 1/( 3.1)²) $]^{1/4$

T = 5800 × $[$ 1.5( 106.9652 ) ( 1/(9.61) $]^{1/4$

T = 5800 × $[$ 16.69592 $]^{1/4$

T = 5800 × 2.02140152

T = 11724.13 K

Therefore, the surface temperature of Regulus A is 11724.13 K

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A 3-kg rock is thrown upward with a force of 200 N at a location where the local gravitational acceleration is 9.79 m/s2 . Deter

expeople1 [14]

Answer: $56.87m/s^{2}$

Explanation:

If we make an analysis of the net force $F_{net}$ of the rock that was thrown upwards, we will have the following:

$F_{net}=F_{up}-W$ (1)

Where:

$F_{up}=200N$ is the force with which the rock was thrown

$W$ is the weight of the rock

Being the weight the relation between the mass $m=3kg$ of the rock and the acceleration due gravity $g=9.79m/s^{2}$ :

$W=m.g=(3kg)(9.79m/s^{2})$ (2)

$W=29.37 N$ (3)

Substituting (3) in (1):

$F_{net}=200N-29.37 N$ (4)

$F_{net}=170.63 N$ (5) This is the net Force on the rock

On the other hand, we know this force is equal to the multiplication of the mass with the acceleration, according to Newton's 2nd Law:

$F_{net}=m.a$ (6)

Finding the acceleration $a$ :

$a=\frac{F_{net}}{m}$ (7)

$a=\frac{170.63 N}{3kg}$ (8)

Finally:

$a=56.87m/s^{2}$

3 0

2 years ago

A thermally insulated vessel containing a gas whose molar mass is equal to M and the ratio of specific heats cP /cV = γ moves wi

Vika [28.1K]

Answer:

∆T = Mv^2Y/2Cp

Explanation:

Formula for Kinetic energy of the vessel = 1/2mv^2

Increase in internal energy Δu = nCVΔT

where n is the number of moles of the gas in vessel.

When the vessel is to stop suddenly, its kinetic energy will be used to increase the temperature of the gas

We say

1/2mv^2 = ∆u

1/2mv^2 = nCv∆T

Since n = m/M

1/2mv^2 = mCv∆T/M

Making ∆T subject of the formula we have

∆T = Mv^2/2Cv

Multiple the RHS by Cp/Cp

∆T = Mv^2/2Cv *Cp/Cp

Since Y = Cp/CV

∆T = Mv^2Y/2Cp k

Since CV = R/Y - 1

We could also have

∆T = Mv^2(Y - 1)/2R k

6 0

3 years ago

An ambulance with a siren emitting a whine at 1790 Hz overtakes and passes a cyclist pedaling a bike at 2.36 m/s. After being pa

Deffense [45]

Answer:

The speed of the ambulance is 4.30 m/s

Explanation:

Given:

Frequency of the ambulance, f = 1790 Hz

Frequency at the cyclist, f' = 1780 Hz

Speed of the cyclist, v₀ = 2.36 m/s

let the velocity of the ambulance be 'vₓ'

Now,

the Doppler effect is given as:

$f'=f\frac{v\pm v_o}{v\pm v_x}$

where, v is the speed of sound

since the ambulance is moving towards the cyclist. thus, the sign will be positive

thus,

$v_x=\frac{f}{f'}(v+v_o)-v$

on substituting the values, we get

$v_x=\frac{1790}{1780}(343+2.36)-343$

or

vₓ = 4.30 m/s

Hence, <u>the speed of the ambulance is 4.30 m/s</u>

6 0

2 years ago

SCIENCE HELP! ASAP :)

Vinvika [58]

I believe it's D. The centripetal force moves away from the center, but the marble stays there due to the string.

6 0

3 years ago

A guitarist finds that the pitch of one of her strings is slightly flat—the frequency is a bit too low. Should she increase or d

Yuri [45]

Answer:

The guitarist should increase the tension of the string.

Explanation:

The frequency of a vibrating string is determined by fn=(n/(2L))√T/μ. So if the tension in the string increased, the rate at which it vibrates (the frequency) will also increase.

Therefore it is advisable that she increase the tension of the string.

I hope it helps, please give brainliest if it does

6 0

3 years ago