Au-shaped tube, open to the air on both ends, contains mercury. water is poured into the left arm until the water column is 10.0 cm deep. how far upward from its initial position does the mercury in the right arm rise?

0.368 cm

Explanation:

x = distance by which the mercury rise

d = depth of the water = 10 cm = 0.10 m

ρ = density of water = 1000 kgm⁻³

ρ' = density of mercury = 13600 kgm⁻³

P₀ = atmospheric pressure

Using equilibrium of pressure on both side

P₀ + ρ g d = P₀ + ρ' g (2x)

(1000) (0.10) = (13600) (2x)

x = 0.00368 m

x = 0.368 cm

i would think it's food chain

here we know that ball will drop down by total distance 3.2 m

so here we know that

now using kinematics

now in the same time it will cover horizontal distance of 0.75 m

so speed of the ball will be