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Mrrafil [7]
3 years ago
8

Assume that all four H-R diagrams below represent a star in different stages of its life, after it starts to fuse hydrogen in it

s core. Rank the HR diagrams based on when each stage occurs, from first to last.
Physics
2 answers:
Mumz [18]3 years ago
3 0
The right answer for the question that is being asked and shown above is that: "1-4-3-2." (main sequence-->red giant-->supergiant-->white dwarf). Assume that all four H-R diagrams below represent a star in different stages of its life, after it starts to fuse hydrogen in its core.


Ilia_Sergeevich [38]3 years ago
3 0

Answer and explanation;

-The diagram at the left represents the Sun (or any other one-solar-mass star) as a hydrogen-burning main-sequence star, with spectral type G and one solar luminosity. The next diagram shows the Sun after it has exhausted its core hydrogen and left the main sequence, making it a sub-giant with energy generated by hydrogen burning in a shell around an inert helium core.

-The third diagram shows the Sun a little later; its energy source is still hydrogen shell burning, but at this point it has expanded in size so much that it is a red giant. The final diagram (far right) shows the white dwarf corpse of a one-solar-mass star; it is hot because it is the exposed core of the dead star, but dim because it is small in size.

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Gekata [30.6K]

Answer:

1. Check Your Wheels

2. Check Your chain for rust

3. Check Your Brakes

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7 0
2 years ago
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An ore sample weighs 17.50 N in air. When the sample is suspended by a light cord and totally immersed in water, the tension in
valkas [14]

Answer:

Volume of the sample: approximately \rm 0.6422 \; L = 6.422 \times 10^{-4} \; m^{3}.

Average density of the sample: approximately \rm 2.77\; g \cdot cm^{3} = 2.778 \times 10^{3}\; kg \cdot m^{3}.

Assumption:

  • \rm g = 9.81\; N \cdot kg^{-1}.
  • \rho(\text{water}) = \rm  1.000\times 10^{3}\; kg \cdot m^{-3}.
  • Volume of the cord is negligible.

Explanation:

<h3>Total volume of the sample</h3>

The size of the buoyant force is equal to \rm 17.50 - 11.20 = 6.30\; N.

That's also equal to the weight (weight, m \cdot g) of water that the object displaces. To find the mass of water displaced from its weight, divide weight with g.

\displaystyle m = \frac{m\cdot g}{g} = \rm \frac{6.30\; N}{9.81\; N \cdot kg^{-1}} \approx 0.642\; kg.

Assume that the density of water is \rho(\text{water}) = \rm  1.000\times 10^{3}\; kg \cdot m^{-3}. To the volume of water displaced from its mass, divide mass with density \rho(\text{water}).

\displaystyle V(\text{water displaced}) = \frac{m}{\rho} = \rm \frac{0.642\; kg}{1.000\times 10^{3}\; kg \cdot m^{-3}} \approx 6.42201 \times 10^{-4}\; m^{3}.

Assume that the volume of the cord is negligible. Since the sample is fully-immersed in water, its volume should be the same as the volume of water it displaces.

V(\text{sample}) = V(\text{water displaced}) \approx \rm 6.422\times 10^{-4}\; m^{3}.

<h3>Average Density of the sample</h3>

Average density is equal to mass over volume.

To find the mass of the sample from its weight, divide with g.

\displaystyle m = \frac{m \cdot g}{g} = \rm \frac{17.50\; N}{9.81\; N \cdot kg^{-1}} \approx 1.78389 \; kg.

The volume of the sample is found in the previous part.

Divide mass with volume to find the average density.

\displaystyle \rho(\text{sample, average}) = \frac{m}{V} = \rm \frac{1.78389\; kg}{6.42201 \times 10^{-4}\; m^{3}} \approx 2.778\; kg \cdot m^{-3}.

3 0
3 years ago
After fixing a flat tire on a bicycle you give the wheel a spin. Its initial angular speed was 5.45 rad/s and it rotated 14.4 re
Komok [63]

Answer:

(a) α = -0.16 rad/s²

(b) t = 33.2 s

Explanation:

(a)

Applying 3rd equation of motion on the circular motion of the tire:

2αθ = ωf² - ωi²

where,

α = angular acceleration = ?

ωf = final angular velocity = 0 rad/s (tire finally stops)

ωi = initial angular velocity = 5.45 rad/s

θ = Angular Displacement = (14.4 rev)(2π rad/1 rev) = 28.8π rad

Therefore,

2(α)(28.8π rad) = (0 rad/s)² - (5.45 rad/s)²

α = -(29.7 rad²/s²)/(57.6π rad)

<u>α = -0.16 rad/s²</u>

<u>Negative sign shows deceleration</u>

<u></u>

(b)

Now, we apply 1st equation of motion:

ωf = ωi + αt

0 rad/s = 5.45 rad/s + (-0.16 rad/s²)t

t = (5.45 rad/s)/(0.16 rad/s²)

<u>t = 33.2 s</u>

6 0
3 years ago
The force F is expressed in terms of the mass “m” and acceleration “a” according to the
LekaFEV [45]

Answer:

F = [M] × [L1 T-2] = M1 L1 T-2.

Explanation:

Therefore, Force is dimensionally represented as M1 L1 T-2.

5 0
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6 0
3 years ago
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