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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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Answer:

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A point charge q1=2.0μC is located on the positive y axis at y=0.30m, and an identical charge q2 is at the origin. Find the magn
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Answer:

(A) 0.279N at angle 38.02°

(B) 0.701N

(C) 14.19°

Explanation:

(A) The net force on q3 is given as:

F = Fxi + Fyj

Fx is the x component of the force

Fy is the y component of the force

Fx = -F(1, 3)cos(90 - x) + F(2, 3)cos0

Fy = -F(2, 3)cosx - F(2, 3)cos90 = -F(2, 3)cosx

First let us find y and angle x from the diagram.

Using Pythagoras theorem,

y² = 0.3² + 0.4²

y² = 0.25

y = 0.5m

Using SOHCAHTOA to find x,

sinx = 0.4/0.5

x = 53.13°

Electrostatic force, F is given as:

F = kqQ/r²

Where k = Coulumbs constant

F(1,3) = (k*q1*q3) / r²

F(1, 3) = (9 * 10^9 * 2.0 * 10^(-6) * 4.0 * 10^(-6)) / (0.5²)

F(1, 3) = 0.288N

F(2,3) = (k*q2*q3) / r²

F(2, 3) = (9 * 10^9 * 2.0 * 10^(-6) * 4.0 * 10^(-6)) / (0.4²)

F(2, 3) = 0.45N

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Fx = -0.288cos36.87 + 0.45

Fx = 0.22N

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Fy = 0.172N

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The magnitude of the force will be

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F(mag) = 0.279N

The direction of the force makes will be

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θ = 38.02° to the x axis.

(B) q2 = - 2.0 * 10^(-6)

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F(2,3) = (k*q2*q3) / r²

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F(mag) = √((-0.68)² + 0.172²)

F(mag) = 0.701N

(C) The direction of the force makes will be

tanθ = 0.172/0.68

θ = 14.19° to the x axis

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