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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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What is the half life-life of your 100 atoms of Carbon-14?
Naddika [18.5K]

Answer:

Answer: It takes 5,730 years for half the carbon-14 to change to nitrogen; this is the half-life of carbon-14. After another 5,730 years only one-quarter of the original carbon-14 will remain

3 0
2 years ago
A body is placed on a rough inclined plane. Why does the frictional force decrease with the increase of angle of inclination
kondor19780726 [428]
The frictional force is directly proportional to the force that is perpendicular on the surface.

When the body is placed on a horizontal level with zero inclination, the only force acting on the body is the gravitational force which always pulls the body down. The gravitational force, in this case, is the perpendicular force to the surface. Accordingly, this entire force is used to generate friction

Now as the inclination of the surface increases, the gravitational force is no longer the perpendicular force of the body, its value decreases, which means only a part is used to generate frictional force. Consequently, frictional force decreases.

When the inclination reaches 90 degrees, the gravitational force does not act along the normal and accordingly, no friction force is generated.

7 0
3 years ago
A uniform plank of mass 10kg and length 10m rests on two supports, A and B as shown. A boy of weight 500N stands at a distance o
kifflom [539]

Answer:

U² = 142.86 N

U¹ = 357.14 N

Explanation:

Taking summation of the moment about point A, we get the following equilibrium equation: (taking clockwise direction as positive)

W(2\ m) - U^2(7\ m) = 0

where,

W = weight of boy = 500 N

U² = reaction ay B = ?

Therefore,

(500\ N)(2\ m)-(U^2)(7\ m)=0\\U^2=\frac{1000\ Nm}{7\ m}\\

<u>U² = 142.86 N</u>

Now, taking summation of forces on the plank. Taking upward direction as positive, for equilibrium position:

W-U^1-U^2=0\\500\ N - 142.86\ N = U^1\\

<u>U¹ = 357.14 N</u>

3 0
3 years ago
A particle with charge 5 . 0- µ C is placed at the corner of a cube. (Physics Help)? A particle with charge 5.0-µC is placed at
Svetradugi [14.3K]

Gauss law states that the electric flux through any closed surface is proportional to the net electric charge inside the surface. This is expressed mathematically in the form of:

Φ = Q / εo

Where,

Φ = the electric flux = unknown (which we have to find for)

Q = the net electric charge = 5.0 µC = 5 E-6 C

εo = the permittivity of free space = a constant value = 8.85 E-12 C^2 / N m^2

Plugging in the values into the equation will result in:

Φ = 5 E-6 C / (8.85 E-12 C^2 / N m^2)

Φ = 564,971.75 Wb = <span>5.6 x 10^5 Wb </span>

6 0
3 years ago
At what height h above the ground does the projectile have a speed of 0.5v?
dedylja [7]

Answer:height above ground at which projectile have velocity

0.5v is (0.0375v^2)

Explanation:

Using Vf = Vi - gt

Where Vf is final velocity

Vi is initial velocity

g is the acceleration due to gravity

t is the time taken

So, 0.5v = v - gt

t = 0.05v

Therefore height h = vt - 0.5gt^2

Subtitute t

h = 0.05v^2 - 0.0125v^2

h = 0.0375v^2

3 0
3 years ago
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