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aivan3 [116]
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?
Engineering
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

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Answer:

motion ------> electrical. winds push the turbines which generate a magnetic fields which in turn, generates electricity

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3 years ago
Select the correct answer.
Ulleksa [173]
The answer is A. Immediately inform her colleague
4 0
3 years ago
Air is contained in a vertical piston–cylinder assembly such that the piston is in static equilibrium. The atmosphere exerts a p
oee [108]

Answer:

a) 24 kg

b) 32 kg

Explanation:

The gauge pressure is of the gas is equal to the weight of the piston divided by its area:

p = P / A

p = m * g / (π/4 * d^2)

Rearranging

p * (π/4 * d^2) = m * g

m = p * (π/4 * d^2) / g

m = 1200 * (π/4 * 0.5^2) / 9.81 = 24 kg

After the weight is added the gauge pressure is 2.8kPa

The mass of piston plus addded weight is

m2 = 2800 * (π/4 * 0.5^2) / 9.81 = 56 kg

56 - 24 = 32 kg

The mass of the added weight is 32 kg.

5 0
3 years ago
If the head loss in a 30 m of length of a 75-mm-diameter pipe is 7.6 m for a given flow rate of water, what is the total drag fo
Stolb23 [73]

Answer:

526.5 KN

Explanation:

The total head loss in a pipe is a sum of pressure head, kinetic energy head and potential energy head.

But the pipe is assumed to be horizontal and the velocity through the pipe is constant, Hence the head loss is just pressure head.

h = (P₁/ρg) - (P₂/ρg) = (P₁ - P₂)/ρg

where ρ = density of the fluid and g = acceleration due to gravity

h = ΔP/ρg

ΔP = ρgh = 1000 × 9.8 × 7.6 = 74480 Pa

Drag force over the length of the pipe = Dynamic pressure drop over the length of the pipe × Area of the pipe that the fluid is in contact with

Dynamic pressure drop over the length of the pipe = ΔP = 74480 Pa

Area of the pipe that the fluid is in contact with = 2πrL = 2π × (0.075/2) × 30 = 7.069 m²

Drag Force = 74480 × 7.069 = 526468.1 N = 526.5 KN

3 0
3 years ago
A steel bar 100 mm (4.0 in.) long and having a square cross section 20 mm (0.8 in.) on an edge is pulled intension with a load o
grigory [225]

Answer:

The elastic modulus of the steel is 139062.5 N/in^2

Explanation:

Elastic modulus = stress ÷ strain

Load = 89,000 N

Area of square cross section of the steel bar = (0.8 in)^2 = 0.64 in^2

Stress = load/area = 89,000/0.64 = 139.0625 N/in^2

Length of steel bar = 4 in

Extension = 4×10^-3 in

Strain = extension/length = 4×10^-3/4 = 1×10^-3

Elastic modulus = 139.0625 N/in^2 ÷ 1×10^-3 = 139062.5 N/in^2

7 0
3 years ago
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