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sammy [17]
3 years ago
11

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?A. The surface will feel a stronger gravitational force during the AGB phase because the star is larger.
B. The surface will feel a weaker gravitational force during the AGB phase because it is farther from the center of the star.
C. The surface will feel the same gravitational force during both phases.
Physics
1 answer:
kiruha [24]3 years ago
6 0

Answer:

Option B

Explanation:

Gravitational force is a force that attracts two bodies (with a mass) towards each other. If an object has a higher mass, the gravitational pull will be greater.

According to Newton’s inverse square law:

<em>"The gravitational force is inversely proportional to the square of the distance between two bodies."</em>

About this question, the greater the distance between two gravitating bodies, the weaker is the gravitational force between them.

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Answer:

d) 1.2 mT

Explanation:

Here we want to find the magnitude of the magnetic field at a distance of 2.5 mm from the axis of the coaxial cable.

First of all, we observe that:

- The internal cylindrical conductor of radius 2 mm can be treated as a conductive wire placed at the axis of the cable, since here we are analyzing the field outside the radius of the conductor. The current flowing in this conductor is

I = 15 A

- The external conductor, of radius between 3 mm and 3.5 mm, does not contribute to the field at r = 2.5 mm, since 2.5 mm is situated before the inner shell of the conductor (at 3 mm).

Therefore, the net magnetic field is just given by the internal conductor. The magnetic field produced by a wire is given by

B=\frac{\mu_0 I}{2\pi r}

where

\mu_0 is the vacuum permeability

I = 15 A is the current in the conductor

r = 2.5 mm = 0.0025 m is the distance from the axis at which we want to calculate the field

Substituting, we find:

B=\frac{(4\pi\cdot 10^{-7})(15)}{2\pi(0.0025)}=1.2\cdot 10^{-3}T = 1.2 mT

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Answer:

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