6) two small balls, a and b, attract each other gravitationally with a force of magnitude f. if we now double both masses and the separation of the balls, what will now be the magnitude of the attractive force on each one?

The magnitude of the attractive force is unchanged

Explanation:

Newton's Law of Universal Gravitation: The law states that the force of attraction between two mass m₁ and m₂ is directly proportional to the product of the masses and inversely proportional to the square of the distance (d) between them.

Mathematically it is represented as

F = Gm₁m₂/d₂ Equation 1

Where G = universal Constant, F = Force of attraction or repulsion, m₁ = mass of the first body,  m₂ = mass of the second body, d = distance between the masses.

When both masses and the separation of the ball are doubled,

I.e

Fₙ = G(2m₁)(2m₂)/(2d)²

Where Fₙ = the new gravitational force, when both masses and distance are separation are doubled.

Fₙ = 4Gm₁m₂/4d²

Fₙ = Gm₁m₂/d² Equation 2

Comparing Equation 1 and equation 2,

Fₙ = F

Therefore the magnitude of the attractive force between the balls when their masses and separation are doubled remains the same.

I.e The magnitude of the attractive force is unchanged,

power out to power in less than 100%

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft