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solmaris [256]
3 years ago
10

A ball is tossed straight up from the surface of a small, sphericalasteroid with no atmosphere. The aball rises to a height equa

l tothe asteroid's raidus and then falls straight down toward thesurface of the asteroid.
1. What forces act on the ball while it is on the way up?

1) only a decreasing gravitational force that acts downward.
2) only an increasing gravitational forces that acts downward
3) only a constant gravitation force that acts downward
4) both a constant gravitational force that acts downward and adecreasing force that acts upward
5) no forces act on the ball

2. the acceleration of the ball at the top of its path is1) at its max vlue for the ball's flightb) Equal to the accleration at the surface of the asteroidc) equal to .5 the acceleration at the surface of the asteroidd) equal to 1/4 the acceleration at the surface of theasteroide) zero
Physics
1 answer:
Anastaziya [24]3 years ago
4 0

Answer:

1. 3) only a constant gravitation force that acts downward

2. b) Equal to the accleration at the surface of the asteroidc)

Explanation:

1.

  • In absence of atmosphere, the force that will act on the steroid is the force of gravity due to asteroid.
  • Gravitational force is a long range force acting always attractive in nature.
  • It is a contact-less field force which does not requires a medium.

Mathematically given as:

F=G.\frac{M_1.M_2}{R^2}

where:

G = gravitational constant

M_1\ \&\ M_2 are the mass of the two objects

R= radial distance between the two objects.

2.

The acceleration of the ball at the top height of the path is still under the influence of the gravity of the two masses so it will be equal to the acceleration due to gravity at the surface of the asteroid. Acceleration always remains constant.

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((1/2)mv²+m*g*h)final = ((1/2)mv²+m*g*h)initial , We divided by m both sides of the equation:

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(1/2) (vB)² + (9.8)*(14.7) =  0 + (9.8)(26.8 )

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