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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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When a mass of 0.350 kg is attached to a vertical spring and lowered slowly, the spring stretches 12.0 cm. The mass is now displ
Ray Of Light [21]

Answer:

period of oscillations is 0.695 second

Explanation:

given data

mass m = 0.350 kg

spring stretches x = 12 cm = 0.12 m

to find out

period of oscillations

solution

we know here that force

force = k × x   .........1

so force = mg =  0.35 (9.8)  = 3.43 N

3.43 = k × 0.12

k = 28.58 N/m

so period of oscillations is

period of oscillations = 2π × \sqrt{\frac{m}{k} }   ................2

put here value

period of oscillations = 2π × \sqrt{\frac{0.35}{28.58} }  

period of oscillations = 0.6953

so period of oscillations is 0.695 second

4 0
3 years ago
The coefficient of static friction between a 3.00 kg crate and the 35.0o incline is 0.300. What minimum force F must be applied
Yakvenalex [24]

Answer:

So the minimum force is

32.2Newton

Explanation:

To solve for the minimum force, let us assume it to be F (N)

So

F=mgsinA

But

=>>>> coefficient of static friction x (F + mgcosA

=>3 x 9.8 x sin35 = 0.3 x (F + 3 x 9.8 x cos35)

So making F subject of formula

F + 24.0 = 56.2

F = 32.2N

3 0
3 years ago
0/2 File Limit
slamgirl [31]

Answer:

Speed at which it will reach the ground is given as

v_f = 46.8 m/s

Total time for which it will remain in air is given as

t = 6.3 s

Explanation:

As we know that the object is projected upwards with speed

v_i = 15 m/s

g = - 9.81 m/s^2

now when it will reach the ground then we have

y = v_y t + \frac{1}{2} at^2

so we have

-100 = 15 t - \frac{1}{2}(-9.81) t^2

4.905 t^2 - 15 t - 100 = 0

so we have

t = 6.3 s

Now speed of the object when it reaches the ground is given as

v_f = v_i + at

v_f = -15 + (9.81)(6.3)

v_f = 46.8 m/s

8 0
3 years ago
Two stones are launched from the top of a tall building. One stone is thrown in a direction 25.0cm above the horizontal with a s
fomenos

Answer:

Explanation:

the one thrown below the horizontal is going straight down, while the one above the horizontal will experience a projectile motion which will makes it move farther away from the building where it was projected.

6 0
3 years ago
A boy throws a baseball onto a roof and it rolls back down and off the roof with a speed of 3.75 m/s. If the roof is pitched at
dalvyx [7]

Answer:

a) t = 0.528 s

b) D = 1.62 m

Explanation:

given,

speed of the baseball = 3.75 m/s

angle made with the horizontal = 35°

height of the roof edge = 2.5 m

using equation of motion

s = ut +\dfrac{1}{2}gt^2

2.5 = vsin\theta \ t +\dfrac{1}{2}gt^2

2.5 = 3.75\ sin35^0 \ t +\dfrac{1}{2}\times 9.8 t^2

4.9 t² + 2.15 t - 2.5 = 0

on solving the above equation

t = 0.528 s

b) D = v cos θ × t

D = 3.75 × cos 35° ×0.528

D = 1.62 m

3 0
3 years ago
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