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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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Which of the following is an example of heat transfer by convection?
erastovalidia [21]

Answer: A. water within a pan is heated and flows in a circular path.

Explanation: Convection is a type of heat transfer that occurs in liquids and gases. Convection occurs when there is a contrast in temperature between two parts of a liquid or gas. The hot part of a fluid rises above, and the cooler part sinks to the bottom resulting to circular motion of the fluid.

6 0
4 years ago
Later research indicated that electric current is actually the flow of ____
iren [92.7K]

Answer:

Electrons.

Explanation:

Electricity was discovered before the discovery of electrons by J.J Thompson in 1896. Before the electron, it was thought that it is the positive ions that move through the wire and carry current—that's why today the conventional current represents the flow of positive charges.

After J.J Thompson's discovery of the electrons, it was realized that it is the electrons that actually carry the current through the conductor. But changing the direction of the conventional current didn't seem appropriate, and that's why the convention continues to be used to this day—reminding us that once it were the positive ions that were thought to carry the current.

6 0
3 years ago
Suppose you wanted to use the Earth's magnetic field to make an AC generator at a location where the magnitude of the field is 0
Likurg_2 [28]

Answer:

The angular velocity I would have to rotate it in order to generate an emf of amplitude 1.0 V is 254.65 rad/s

Explanation:

given information:

B = 0.5 mT = 0.0005 T

N = 1000

r = 5 cm = 0.05 m

emf, ε = 1 V

according to Faraday's law

ε = -N dΦ/dt, Φ = B A

  = - N d( B A)/dt

  = - N d( B A cos ωt)/dt

   = - N B A d(cos ωt)/dt

   = N B A ω sin ωt

A = πr², so

ε = N B πr² ω sin ωt

where

ε = emf

N = number of coil turn

B = magnetic field

r = radius

ω = angular velocity

Φ = magnetic flux

emf maximum, sin ωt = 1. So,

ε = N B πr² ω

ω = ε/N B πr²

   = 1/[(1000) (0.0005) π (0.05)²

   = 254.65 rad/s

6 0
3 years ago
What is the iodide ion concentration in a solution if the addition of an excess of 0.100 m pb(no3)2 to 42.9 ml of the solution p
Katen [24]
Moles Pbl2 =  0.8628 g  :  461.01 g/mol   = 0.001871

moles I = 2 x  0.001871 = 0.003742

[I-] = 0.003742/ 0.0429

= 0.0872 M
3 0
3 years ago
Anybody know the answer??
omeli [17]

Answer:

Centimeter

Explanation:

We know the measurement is defined as:

1 meter= 100cm

1 cm= 1/100 meter

1 centimeter= 0.01 meter

Therefore, 0.01 meter is equal to 1 cm or 1 centimeter.

6 0
3 years ago
Read 2 more answers
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