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irinina [24]
3 years ago
15

How will the gravitational force on a piece of the surface of the star (m1) by the mass of the rest of the star (m2) (effectivel

y located at a point at the center of the star) compare between the AGB and main-sequence phases of a particular star, assuming its mass stays the same?
Physics
2 answers:
NikAS [45]3 years ago
6 0

Answer: The surface will feel a weaker gravitational force during The AGB phase because it farther from the center of the star.

Explanation: The surface will feel a weaker gravitational force during the AGB phase because it farther from the center of the star. Surface mass loss via a wind during the AGB phase and/or via catastrophic ... of heat, pressure, and gravitational diffusion are ignored.

Anna [14]3 years ago
6 0

Answer:

Answer: Gravitational weak force will be felt on the surface during the asymptotic giant branch phase because it is far away from the middle of the star.

Explanation:

(AGB) which is know as asymptotic giant branch is a space on the Hertzsprung–Russell drawing which been populated by cool luminous stars which are evolved.

When the surface will be felt to be weak, is during the period of asymptotic giant branch(AGB) phase, which will be very far away from the middle of the star.

The will be loss of the surface mass through a wind during the AGB phase and/or through catastrophic of gravitational diffusion, pressure, heat are been left out.

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Lera25 [3.4K]
Around 750 to 1,500.

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5 0
4 years ago
In preparation for the final exam, our astronomy study group has reconvened to discuss Mercury's unique orbital properties. They
grigory [225]

Answer:

The only incorrect statement is from student B

Explanation:

The planet mercury has a period of revolution of 58.7 Earth days and a rotation period around the sun of 87 days 23 ha, approximately 88 Earth days.

Let's examine student claims using these rotation periods

Student A. The time for 4 turns around the sun is

           t = 4 88

           t = 352 / 58.7 Earth days

In this time I make as many rotations on itself each one with a time to = 58.7 Earth days

           #_rotaciones = t / to

           #_rotations = 352 / 58.7

           #_rotations = 6

therefore this statement is TRUE

student B. the planet rotates 6 times around the Sun

          t = 6 88

          t = 528 s

The number of rotations on itself is

           #_rotaciones = t / to

           #_rotations = 528 / 58.7

           #_rotations = 9

False, turn 9 times

Student C. 8 turns around the sun

           t = 8 88

           t = 704 days

the number of turns on itself is

            #_rotaciones = t / to

            #_rotations = 704 / 58.7

            #_rotations = 12

True

The only incorrect statement is from student B

6 0
2 years ago
QUESTION 1
Molodets [167]
Question 1: C Question 2: B, Hope this Helps!
3 0
3 years ago
6. Show that the weight of an object on the moon is 1/6 its weight on earth.​
mojhsa [17]

Taking ratio of W & w. ≈ 6 . w = 1/6 W. Therefore , Weight of an object on the moon is 1/6 of its weight on the earth.

5 0
3 years ago
A 1.5m wire carries a 6 A current when a potential difference of 68 V is applied. What is the resistance of the wire?
ANTONII [103]

Answer:

11.3 \Omega

Explanation:

We can find the resistance of the wire by using Ohm's law:

V=RI

where

V is the voltage applied

R is the resistance

I is the current

In this problem, we know I = 6 A and V = 68 V, so we can re-arrange the equation to find the resistance of the wire:

R=\frac{V}{I}=\frac{68 V}{6 A}=11.3 \Omega

6 0
3 years ago
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