Answer::
a) 1,231,504 N b) 318,322 N. c) 493.33 kPa.
Explanation:
a) We can calculate the weight of a cylindrical colum of water, as follows:
Fg = δH2O * V * g = δH2O * π* r2 * h * g = 1,000 kg/m3*π*(1.00)2 m2 *40.0 m* 9.8 m/s2
Fg = 1,231,504 N.
b) In order to get the force exerted by air on a disk at the water´s surface, we can apply the definition of pressure, and use the value for atmospheric pressure, as follows:
Fa = Pa.A = 101.325 k N/m2 * π* r2 = 101.325 kN/m2 * π * (1.00) m2
Fa = 318,322 N
c) At a depth of 40.0 m, the absolute pressure supported by the water column is the sum of the atmospheric pressure, plus the pressure created by the weight of the water column, as follows:
p = po + δH2O * g * h
p= 101.325 kN/m2 + 1,000 kg/m3 * 9.8 m/s2 * 40.0 m = 101,325 N/m2 + 392,000 N/m2
p= 493,325 N/m2 = 493.33 kPa.
(a) The weight of the water is 1.23×10⁶ N
(b) The force exerted by the air on the disk is 3.14×10⁵ N
(c) The pressure at a depth of 40.0 m is 4.93×10⁵ Pa
(a) The mass of water is given by:
m = ρV
where ρ = 1000 kg/m³ is the density of water
V is the volume
V = πr²h
V = 3.14×(1)²×40
V = 125.6 m³
thus,
m = 1000×125.6 m³
m = 125600 kg
Now the weight of water:
W = mg = 126500×9.8 N
W = 1.23×10⁶ N
(b) At the water surface, the pressure is equal to atmospheric pressure that is:
Pₐ = 1.01×10⁵ Pa
The force is given by;
F = PA
F = 1.01×10⁵×πr²
F = 1.01×10⁵×3.14×(1)²
F = 3.14×10⁵ N
(c) The pressure at the depth h is given by:
P = Pₐ + ρgh
P = 1.01×10⁵ + 1000×9.8×40
P = 4.93×10⁵ Pa
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