3. If the gravitational force between 2 objects is 50 N, what is the gravitational force when the distance between the two is increased to four times the distance?

The gravitational  force : 3.125 N

Given

F₁ = 50 N

Required

F₂

Solution

Newton's Gravity Law:  

with F = gravitational force,  

G = gravitational constant,  

m1, m2 = mass of object,  

r = distance between two objects.  

The value of m and G are the same , so :

F₁ ≈ 1/r₁²

The distance between the two is increased to four times:

r₂ = 4r₁

F₂ = 1/(4r₁)²

F₂ = 1/16r₁²

F₂ = F₁ x 1/16

F₂ = 50 : 16

F₂ = 3.125 N

waves moving from a fast medium to a slow medium will experience a reflection as light travels more slowly than red light. total internal reflection occurs when the critical angle is.

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