gravitational force between two objects is given as
F = G m₁ m₂/r²
where m₁ = mass of first object , m₂ = mass of second object , r = distance between the two objects .
Initial case :
m₁ = m₂ = m
gravitational force between the objects is given as
F = G m²/r²
Final Case :
m₁ = m₂ = 3 m
new gravitational force between the objects is given as
F' = G (3m)²/r²
F' = 9 G m²/r²
F' = 9 F
hence the gravitational force between the two objects becomes 9 times.
Answer:
Explanation:
We shall apply conservation of momentum law in vector form to solve the problem .
Initial momentum = 0
momentum of 12 g piece
= .012 x 37 i since it moves along x axis .
= .444 i
momentum of 22 g
= .022 x 34 j
= .748 j
Let momentum of third piece = p
total momentum
= p + .444 i + .748 j
so
applying conservation law of momentum
p + .444 i + .748 j = 0
p = - .444 i - .748 j
magnitude of p
= √ ( .444² + .748² )
= .87 kg m /s
mass of third piece = 58 - ( 12 + 22 )
= 24 g = .024 kg
if v be its velocity
.024 v = .87
v = 36.25 m / s .
Answer:
Since the net force is to the right (in the direction of the applied force), then the applied force must be greater than the friction force. The friction force can be determined using an understanding of net force as the vector sum of all the forces.
Explanation:
I’m pretty sure you times them so 1 with A, 2 with e, 3 with C, and 4 with B
