Question

The mattress of a water bed is 2.00 m long by 2.00 m wide and 30.0 cm deep. Find the weight of the water in the mattress. Knowing that the density of water rho = 1000 kg/m3, find the pressure exerted by the water on the floor when the water bed rests in its normal position. Assume the entire lower surface of the bed makes contact with the floor.

Answer 1

Answer:

2943 Pa

Explanation:

$\rho$ = Density of water = $1000\ kg/m^3$

g = Acceleration due to gravity = $9.81\ m/s^2$

V = Volume

A = Area

Mass of the blanket

$m=\rho V\\\Rightarrow m=1000\times 2\times 2\times 0.3\\\Rightarrow m=1200\ kg$

Force on the ground applied due to its own weight

$F=mg\\\Rightarrow F=1200\times 9.81\\\Rightarrow F=11772\ N$

Pressure is given by

$P=\dfrac{F}{A}\\\Rightarrow P=\dfrac{11772}{2\times 2}\\\Rightarrow P=2943\ Pa$

The pressure exerted by the water on the floor  is 2943 Pa

Answer 2

Answer:

2940 Pascal

Explanation:

Length, l = 2 m

width, w = 2 m

depth, h = 30 cm = 0.3 m

density of water, ρ = 1000 kg/m³

Let the pressure at the depth is P.

P = h x ρ x g

P = 0.3 x 1000 x 9.8

P = 2940 Pascal

Thus, the pressure at the depth is 2940 Pa.