8,000,000 kg of water are at the verge of dropping over Niagara Falls to the rocks 50.0 meters below. What is the change in grav
1 answer:
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Answer:
132.9 cm
Explanation:
Data provided in the question:
Radius of the basll = 76 cm = 0.76 m
Side of the box = 200 cm = 2 m
Density of the ball and cube are equal
let the density be 'D'
Now,
Mass of ball, M = Volume × Density
= × D
= × D
= 1.838D
Mass of cube, m = L³ × D
= 2³ × D
= 8D
Thus,
center of mass, y = [ My₁ + my₂ ] ÷ [M + m]
here,
y₁ = center of mass of ball with respect to floor
as the center mass of sphere lies at the center of the sphere
= Length of cube + radius of sphere
= 2 + 0.76
= 2.76 m
y₂ = Center of mass of cube = = 1 m
Thus,
y = [ ( 1.838D × 2.76 ) + (8D × 1 ) ] ÷ [1.838D + 8D]
= 13.07288D ÷ 9.838D
= 1.329 m
or
= 132.9 cm