Question

A long, thin metal tube of cross-sectional area 1.60 cm2 is sealed into the top of a hollow cube of edge length 0.150 m. Water fills the cube and extends upward a distance h into the tube. Find h if the water exerts a force of 1.00 103 N on the base of the cube. What is the weight of the water?

Answer 1

Step-by-step explanation:

Given that:

The area of the metal tube = 1.60 cm² = 1.60 × 10⁻⁴ m²

Length of cube L = 0.150 m

The exerted force by water = 1.00 × 10³ N

At the top of the cube;

The force on water is F = pgL

F = 9.8 m/s² × (1000 kg/m³) (0.150 m)(0.150 m)²

F = 33.075 N

∴

`pgh = (F)/(A)`

`h = (F)/(Apg)`

`h = (33.075)/((1.60 * 10^(-4) \ m^2) * (1000 \ kg/m^3)(9.8))`

h = 21.09 m

The volume of the water V = L³

V = (0.150 m)³

V = 0.003375 m³

weight of the water w = p(V + Ah)

w = 1000 (0.003375 + (1.60 × 10⁻⁴ × 21.09))

w = 6.75 kg