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3 years ago

For main sequence stars what tends to happen to the absolute brightness of a star as the temperature rises? How do you know?

Physics

1 answer:

professor190 [17]3 years ago

6 0

With the increase in the temperature of the star, the brightness of the stars will also increase.

<u>Explanation:</u>

The brightness and surface temperature of stars ordinarily increment with age. A star stays close to its underlying situation on the fundamental arrangement until a lot of hydrogen in the center has been devoured, at that point starts to advance into a progressively brilliant star.

The brightness of a star relies upon its structure and how far it is from the planet. Space experts characterize star brilliance as far as clear extent — how splendid the star shows up from Earth — and outright greatness — how brilliant the star shows up at a standard separation

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Which of the following types of weather data is collected in inches or centimeters?

aksik [14]

The correct answer is C) Rainfall

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3 years ago

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An electrician finds that a 1 m length of a certain type of wire has a resistance of 0.24 Ω . What is the total resistance of th

zlopas [31]

The resistance of a given conductor depends on its electrical resistivity ( $\rho$ ), its length(L) and its cross-sectional area (A), as follows:

$R=\frac{\rho L}{A}$

In this case, we have $L'=138L$ , $\rho'=\rho$ and $A'=A$ . So, the total resistance of the wire with length of 138m is:

$R'=\frac{\rho' L'}{A'}\\R'=\frac{\rho 138L}{A}\\R'=138\frac{\rho L}{A}\\R'=138R\\R'=138(0.24\Omega)\\R'=33.12\Omega$

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3 years ago

Which of the following is not a reason why plants need water? *

solong [7]

Answer:

Explanation:

the plant cools itself down by allowing water to evaporate from their leaves so it doesn't need water to cool down

5 0

3 years ago

A piece of steel is 11.5cm long at 22C. It is heated to 1221C, close to its melting point. How long is it, in cm, at the high te

Nataly [62]

Answer:

The length at the final temperature is 11.7 cm.

Explanation:

We need to use the thermal expansion equation:

$\Delta L=\alpha L_{0}\Delta T$

Where:

L(0) is the initial length
ΔT is the differential temperature, final temperature minus initial temperature (T(f)-T(0))
ΔL is the final length minus the initial length (L(f)-L(0))
α is the coefficient of linear expantion of steel (12.5*10⁻⁶ 1/°C)

So, we have:

$L_{f}-L_{0}=\alpha L_{0}(T_{f}-T_{0})$

$L_{f}=L_{0}+\alpha L_{0}(T_{f}-T_{0})$

$L_{f}=0.115+(12.5*10^{-6})(0.115)(1221-22)$

$L_{f}=0.117\: m$

Therefore, the length at the final temperature is 11.7 cm.

I hope it helps you!

7 0

3 years ago

A cue ball, moving with 9.0 N·s of momentum strikes the nine-ball at rest. The nine-ball moves off with 2.0 N·s in the original

lisov135 [29]

Answer:

P = 7.28 N.s

Explanation:

given,

initial momentum of cue ball in x- direction,P₁ = 9 N.s

momentum of nine ball in x- direction, P₂ = 2 N.s

momentum in perpendicular direction i.e. y - direction,P'₂ = 2 N.s

momentum of the cue after collision = ?

using conservation of momentum

in x- direction

P₁ + p = x + P₂

p is the initial momentum of the nine balls which is equal to zero.

9 + 0 = x + 2

x = 7 N.s

momentum in x-direction.

equating along y-direction

P'₁ + p = y + P'₂

0 + 0 = y + 2

y = -2 N.s

the momentum of the cue ball after collision is equal to resultant of the momentum .

$P = \sqrt{x^2+y^2}$

$P = \sqrt{7^2+(-2)^2}$

P = 7.28 N.s

the momentum of the cue ball after collision is equal to P = 7.28 N.s

7 0

3 years ago