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Vitek1552 [10]
4 years ago
11

The mass luminosity relation L  M 3.5 describes the mathematical relationship between luminosity and mass for main sequence sta

rs. It describes how a star with a mass of 2 M⊙ would have a luminosity of _____________ L⊙ while a star with luminosity of 3,160 L⊙ would have an approximate mass of ________________ M⊙.
Physics
1 answer:
ivanzaharov [21]4 years ago
3 0

Answer:

(a) <u>11.3 L</u>

(b) <u>10 M</u>

Explanation:

The mass-luminosity relationship states that:

Luminosity ∝ Mass^3.5

Luminosity = (Constant)(Mass)^3.5

So, in order to find the values of luminosity or mass of different stars, we take the luminosity or mass of sun as reference. Therefore, write the equation for a star and Sun, and divide them to get:

Luminosity of a star/L = (Mass of Star/M)^3.5 ______ eqn(1)

where,

L = Luminosity of Sun

M = mass of Sun

(a)

It is given that:

Mass of Star = 2M

Therefore, eqn (1) implies that:

Luminosity of star/L = (2M/M)^3.5

Luminosity of Star = (2)^3.5 L

<u>Luminosity of Star =  11.3 L</u>

(b)

It is given that:

Luminosity of star = 3160 L

Therefore, eqn (1) implies that:

3160L/L = (Mass of Star/M)^3.5

taking ln on both sides:

ln (3160) = 3.5 ln(Mass of Star/M)

8.0583/3.5 = ln(Mass of Star/M)

Mass of Star/M = e^2.302

<u>Mass of Star = 10 M</u>

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I attempted to answer and got 0m, please explain how to get to the answer.
mafiozo [28]

The maximum height to which the ball attain before falling back down is 1147.96 m

<h3>Data obtained from the question</h3>

The following data were obtained from the question:

  • Initial velocity (u) = 150 m/s
  • Final velocity (v) = 0 m/s (at maximum height)
  • Acceleration due to gravity (g) = 9.8 m/s²
  • Maximum height (h) =?

<h3>How to determine the maximum height </h3>

The maximum height reached by the ball can be obtained as illustrated below:

v² = u² – 2gh (since the ball is going against gravity)

0² = 150² – (2 × 9.8 × h)

0 = 22500 – 19.6h

Collect like terms

0 – 22500 = –19.6h

–22500 = –19.6h

Divide both side by –19.6

h = –22500 / –19.6

h = 1147.96 m

Thus, the maximum height reached by the ball is 1147.96 m

Learn more about motion under gravity:

brainly.com/question/22719691

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5 0
1 year ago
Two charges that are separated by one meter exert 1-n forces on each other. if the magnitude of each charge is doubled, the forc
Goshia [24]
The electrostatic force between the two charges is
F=k_E  \frac{q_1 q_2}{r^2}
where q1 and q2 are the magnitudes of the two charges, and r the distance between them.

We can see from the formula that F is proportional to the product between the two charges:
F \sim q_1 q_2
so, if the magnitude of each charge is doubled, the new force will get a factor 4:
F' \sim (2 q_1 )(2 q_2 )=4 q_1 q_2 =4 F
So, the new force will be 4 times the original force:
F' = 4 \cdot 1N= 4N
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3 years ago
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3 years ago
Read 2 more answers
There are two identical small metal spheres with charges 38.9 µC and −27.6399 µC. Thedistance between them is 6 cm. The spheres
Greeley [361]

Answer:

2683.3N

Explanation:

According to coulombs law which states that "the force of attraction existing between two charge q1 and q2 is directly proportional to the product of the charges and inversely proportional to the square of the distance (d) between them. Mathematically |F|= k|q1| |q2| /d² where;

F is the force of attraction between the charges

q1 and q2 are the charges

d is the distance between them

k is the coulombs constant

Given |q1|= 38.9 × 10^-6C and |q2| = 27.6399×10^-6C d = 6cm = 0.06m

k = 8.98755 × 109 Nm² /C²

Substituting the given data's in the equation we have;

|F| = 8.98755 × 10^9×38.9×10^-6×27.6399×10^-6/0.06²

|F| = 9.66/0.06²

|F| = 9.66/0.0036

|F| = 2683.3N

The magnitude of the force will be 2683.3N

Note that the modulus of the charges changes negative value of q2 to positive value. The opposite signs of the charges doesn't affect the final calculation, it only tells the force of attraction or repulsion between the charges. Since they are unlike charges, they will attract each other in the field.

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3 years ago
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