Water flows straight down from an open faucet. The cross-sectional area of the faucet is 2.5 × 10-4m2 and the speed of the water is 0.65 m/s as it leaves the faucet. Ignoring air resistance, find the cross-sectional area of the water stream at a point 0.08 m below the faucet.
Here,
initial speed, u = 0.65 m/s
d = 0.08 m
let the final speed is v
Using third euation of motion
v^2 - u^2 =2 a * s
v^2 - 0.65^2 = 2 * 9.8 * 0.08
solving for v
v = 1.41 m/s
Using equation of continuity
A1 * v1 = A2 * v2
0.65 * 2.5 *10^-4 = 1.41 * A2
A2 = 1.15 *10^-4 m^2
the new cross sectional area is 1.15 *10-4 m^2
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