Ask question

Ask question

All categories

2 years ago

5

Estimate the luminosity of a 3 -solar-mass main-sequence star; of a 9 -solar-mass main-sequence star. Can you easily estimate th

e luminosity of a 3 -solar-massred giant star?

1 answer:

Oxana [17]2 years ago

6 0

Answer:

a) 46.76 Lsun

b) 2187 Lsun

c ) You can easily estimate the luminosity of a 3-solar massed giant star because its luminosity is close to the luminosity of the sun

Explanation:

To estimate the luminosity of a celestial body in terms of solar luminosity we use the expression below

= $\frac{L}{Lsun} = (\frac{M}{Msun} )^{3.5}$

a) Estimate the luminosity of a 3 solar mass main sequence star

= $\frac{L}{Lsun} = (\frac{3M}{Msun} )^{3.5}$

Hence ; L(luminosity of a 3 solar mass star ) = 46.76 Lsun

b) Estimate the luminosity of a 9-solar-mass main sequence star

= $\frac{L}{Lsun} = (\frac{9M}{Msun} )^{3.5}$

hence L( luminosity of a 9-solar mass star ) = 2187 Lsun

c ) You can easily estimate the luminosity of a 3-solar massed giant star because its luminosity is close to the luminosity of the sun

You might be interested in

Im passed due someone help meeeeeee

vovangra [49]

Answer:

how are supposed to help when you can't do anything?

8 0

2 years ago

Read 2 more answers

Air at 40C flows over a 2 m long flat plate with a free stream velocity of 7 m/s. Assume the width of the plate (into the paper)

mezya [45]

Complete Question

Air at 40C flows over a 2 m long flat plate with a free stream velocity of 7m/s. Assume the width of the plate (into the paper) is 0.5 m. If the plate is at a co temperature of 100C,find:

The total heat transfer rate from the plate to the air

Answer:

$q=1.7845$

Explanation:

From the question we are told that:

Air Temperature $T_1=40c$

Length $l=2m$

Velocity $v=7m/s$

Width $w=0.5$

Constant temperature $T_t= 100C$

Generally the equation for Total heat Transfer is mathematically given by

$q=hA(T_s-T_\infty)$

Where

h=Convective heat transfer coefficient

$h=29.9075w/m^2k$

Therefore

$q=h(L*B)(T_s-T_\infty)$

$q=29.9075*(2*0.5)(100+273-(40+273))$

$q=1794.45w$

$q=1.7845$

5 0

2 years ago

Determine the dimensions for W if W = P L^3 / (M V^2) where P is a pressure, L is a length, M is a mass, and V is a velocity.

Eva8 [605]

Correct answer is option E. No dimensions

As we know formula Pressure (P) is $\frac{F}{A}$

also,

Dimensional formula of <em>Pressure is </em> $M^{1}L^{-1}T^{-2}$
Dimensional formula of <em>length is L </em>
Dimensional formula of <em>mass is M</em>
Dimensional formula of <em>velocity is </em> $L^{1} T^{-1}$

So, as given W= $\frac{P*L^{3} }{M*V^{2} }$

Dimensional formula of W = $\frac{M^{1}L^{-1}T^{-2} L^{3} }{M^{1} L^{2}T^{-2} }$

since all terms get cancelled

Work is dimensionless i.e no dimensions

Learn more about dimensions here brainly.com/question/20351712

#SPJ10

6 0

1 year ago

BCC lithium has a lattice parameter of 3.5089 3 10–8 cm and contains one vacancy per 200 unit cells. Calculate (a) the number of

Tanya [424]

(a) The number of vacancies per cubic centimeter is 1.157 X 10²⁰

(b) ρ = n X (AM) / v X Nₐ

<u>Explanation:</u>

<u />

Given-

Lattice parameter of Li = 3.5089 X 10⁻⁸ cm

1 vacancy per 200 unit cells

Vacancy per cell = 1/200

(a)

Number of vacancies per cubic cm = ?

Vacancies/cm³ = vacancy per cell / (lattice parameter)³

Vacancies/cm³ = 1 / 200 X (3.5089 X 10⁻⁸cm)³

Vacancies/cm³ = 1.157 X 10²⁰

Therefore, the number of vacancies per cubic centimeter is 1.157 X 10²⁰

(b)

Density is represented by ρ

ρ = n X (AM) / v X Nₐ

where,

Nₐ = Avogadro number

AM = atomic mass

n = number of atoms

v = volume of unit cell

4 0

2 years ago

The amplitudes of the displacement and acceleration of an unbalanced motor were measured to be 0.15 mm and 0.6*g, respectively.

ehidna [41]

Answer:

The speed of shaft is 1891.62 RPM.

Explanation:

given that

Amplitude A= 0.15 mm

Acceleration = 0.6 g

So

we can say that acceleration= 0.6 x 9.81

$acceleration,a=5.88\ \frac{m}{s^2}$

We know that

$a=\omega ^2A$

So now by putting the values

$a=\omega ^2A$

$5.88=\omega ^2 \0.15\times 10^{-3}$

$\omega =198.09\ \frac{rad}{s}$

We know that

ω= 2πN/60

198.0=2πN/60

N=1891.62 RPM

So the speed of shaft is 1891.62 RPM.

4 0

2 years ago