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AVprozaik [17]
3 years ago
13

How are a star's luminosity, mass, and radius related?

Physics
2 answers:
Tatiana [17]3 years ago
8 0

Answer:

The Luminosity of a star depends on BOTH its temperature and its radius (surface area): L is proportional to R2 T4. A hotter star is more luminous than a cooler one of the same radius. A bigger star is more luminous than a smaller one of the same temperature.

Explanation:

hope it helps

help me out by going to my page and answering some questions

9966 [12]3 years ago
7 0

Answer:

The Luminosity of a star depends on BOTH its temperature and its radius (surface area): L is proportional to R2 T4. A hotter star is more luminous than a cooler one of the same radius. A bigger star is more luminous than a smaller one of the same temperature.

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Compare each pair of numbers using <, > or =.
Effectus [21]

Answer:

A. 56 cm < 6 m

B. 7 g > 698 mg

C. 19 g > .021k

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3 years ago
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denis23 [38]

Answer:

nnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnn

Explanation:

3 0
3 years ago
Read 2 more answers
If the net force equals zero, what is true?
Lubov Fominskaja [6]

Answer:

the object is not accelerating.

7 0
3 years ago
A container is filled to a depth of 0.9490.949m with water. On top of the water floats a 1.971.97m-thick layer of oil with densi
Paha777 [63]

Answer:124.14 KPa

Explanation:

Given

Layer of water is 0.949 m thick

Layer of oil is 1.971 m thick with density

\rho _{o}=700 kg/m^3

density of water \rho _{w}=1000 kg/m^3

suppose Atmospheric Pressure be 101.325 KPa

Absolute Pressure at bottom is =Pressure at top +Pressure due to Oil and water

P_{abs}=101.325\times 10^3+\rho _{o}\cdot g\cdot 1.971+\rho _{w}\cdot g\cdot 0.949

P_{abs}=101.325\times 10^3+700\times 9.8\times 1.971+1000\times 9.8\times 0.949

P_{abs}=101.325+13.521+9.3002\ KPa

P_{abs}=124.14 KPa

6 0
4 years ago
A composite wall separates combustion gases at 2400°C from a liquid coolant at 100°C, with gas and liquid-side convection coeffi
evablogger [386]

Answer:

\text{heat loss} = 24864.05 \  W/m^2

Explanation:

If

  • T_1, T_2 are temperatures of gasses and liquid in Kelvins,
  • t_1 and t_2 are thicknesses of gas layer and steel slab in meters,
  • h_1, h_2 are convection coefficients gas and liquid in W/m^2 \cdot K,
  • R_c is the contact resistance in m^2 \cdot K/W,
  • and k_1, k_2 are thermal conductivities of gas and steel in W/m \cdot K,

then: part(a):

\text{heat loss } =  \frac{T_1 - T_2} { \frac{1}{h_1} + \frac{t_1}{t_2} + R_c + \frac{t_2}{k_2} + \frac{1}{h_2}}

using known values:

\text {heat loss} = 2486.05 W/m^2

part(b): Using the rate equation :

\text {heat loss} = h_1 (T_1 - T_{s1})

the surface temperature T_{s1} = 1678.438 \ K

and T_{c1} = T_{s1} - \frac {t_1 (\text{heat loss})}{k_1} = 1664.560 \ K

Similarly

T_{c2} = T_{c1} - R_c (\text{heat loss}) = 421.357 \ K

T_{s2} = T_{c2} - \frac {t_2 (\text{heat loss})}{ k_2} = 397.864 \ K

The temperature distribution is shown in the attached image

3 0
3 years ago
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