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3 years ago

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Physical properties used to measure stars include

2 answers:

mr Goodwill [35]3 years ago

7 0

Answer: Option 'D' is correct.

Explanation:

Physical features of stars are temperature, size, color, brightness, mass.

All are used to measure star.

Basically star is chemical composition which consists brightness, color , mass, size etc.

Hence, Option 'D' is correct.

V125BC [204]3 years ago

5 0

The answer is D. Mass, brightness and size are used to measure the physical properties of the stars. One is the brightness it is a combination of the intrinsic brightness of a star. Second is the size is calculated using the Stephan's Law. And the third is the mass, astronomers use the Stellar mass to describe the mass of the stars.

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The endocrine system consists of: (2 points)

motikmotik

Answer: Glands

Explanation: Glands are the structures of the endocrine system that produce and secrete hormones.

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3 years ago

a 0.0780 kg lemming runs off a 5.36 m high cliff at 4.84 m/s. what is its kinetic energy when it's 2.00 m above the ground

Blababa [14]

Answer:

KE_2 = 3.48J

Explanation:

Conservation of Energy

E_1 = E_2

PE_1+KE_1 = PE_2+KE_2

m*g*h+(1/2)m*v² = m*g*h+(1/2)m*v²

(0.0780kg)*(9.81m/s²)*(5.36m)+(.5)*(0.0780kg)*(4.84m/s)² = (0.0780kg)*(9.81m/s²)*(2m)+KE_2

4.10J+0.914J = 1.53J + KE_2

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KE_2 = 3.48J

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3 years ago

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aleksandr82 [10.1K]

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Using phones is messing with the ozone layer

Explanation:

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3 years ago

The slope of the line on a speed-time graph tells the speed.

iren2701 [21]

From the choices provided, the better answer is ' T ' .

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3 years ago

g In a certain binary-star system, each star has the same mass which is 8.2 times of that of the Sun, and they revolve about the

Mademuasel [1]

To solve this problem it is necessary to apply the concepts related to the Third Law of Kepler.

Kepler's third law tells us that the period is defined as

$T^2 = \frac{4\pi^2 d^3}{2GM}$

The given data are given with respect to known constants, for example the mass of the sun is

$m_s = 1.989*10^{30}$

The radius between the earth and the sun is given by

$r = 149.6*10^9m$

From the mentioned star it is known that this is 8.2 time mass of sun and it is 6.2 times the distance between earth and the sun

Therefore:

$m = 8.2*1.989*10^{30}$

$d = 6.2*149.6*10^6$

Substituting in Kepler's third law:

$T^2 = \frac{4\pi^2 d^3}{2}$

$T^2 = \frac{4\pi^2(6.2*149.6*10^9)^3}{2(6.674*10^{-11} )(8.2*1.989*10^30 )}$

$T=\sqrt{\frac{4\pi^2(6.2*149.6*10^9)^3}{2(6.674*10^{-11} )(8.2*1.989*10^30)}}$

$T = 120290789.7s$

$T = 120290789.7s(\frac{1year}{31536000s})$

$T \approx 3.8143 years$

Therefore the period of this star is 3.8years

7 0

3 years ago