Question

A person is diving in a lake in the depth of h = 9.5 m. The density of the water is rho = 1.0 x103 kg/m3. The pressure of the atmosphere is P0 = 1.0 x 105 Pa. The surface area of the top of the person's head is A = 0.0435 m2.Randomized Variables
h = 7.5 m
A = 0.0385 m2
Part (a) Express the absolute pressure at the depth of h, P, in terms of P0, ?, and h.
Part (b) Calculate the numerical value of P in Pa. P =
Part (c) Express the magnitude of the force exerted by water on the person's head F at the depth h in terms of P and A. 25% Part (d) Calculate the numerical value of F in N.

Answer 1

A) The absolute pressure at the depth h, P in terms of P₀ is $P = P_0+9800h$

B) The numerical value of P in Pa is 193100Pa

C) The magnitude of the force exerted by water on the person's head F at the depth h in terms of P and A is F = PA

D) The numerical value of F in N is 7434.35 N

The depth of the lake, h = 9.5 m

The density of water, $\rho = 1.0 \times 10^3 kg/m^3$

The atmospheric pressure, $P_0= 1.0 \times 10^5 Pa$

The acceleration due to gravity, $g = 9.8 m/s^2$

The absolute pressure is given by the formula:

$P = P_0 + h\rho g$

a) Express the absolute pressure at the depth h, P in terms of P₀ by substituting  $g = 9.8 m/s^2$  and  $\rho = 1.0 \times 10^3 kg/m^3$into the equation above

$P = P_0 + 9.8(10^3)h\\\\P = P_0+9800h$

b) Calculate the numerical value of P in Pa

Substitute  $P_0= 1.0 \times 10^5 Pa$ and h = 9.5 m into the equation in part A

$P = 10^5 + 9800(9.5)\\\\P = 100000 + 93100\\\\P = 193100 Pa$

c) Express the magnitude of the force exerted by water on the person's head F at the depth h in terms of P and A

$Pressure = \frac{Force}{Area} \\P = \frac{F}{A} \\\\F = PA$

d) Calculate the numerical value of F in N

Area, A = 0.0385 m²

F  = 193100(0.0385)

F = 7434.35 N

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Answer 2

a) $\rm P = P_0 + 9800h$  

b) P = 193100 Pa

c) F = PA

d) F = 7434.35 N

Given :

h = 9.5 m

$\rm \rho = 1 \times 10^3\;kg/m^3$

$\rm P_0 = 1\times 10^5 \;Pa$

$\rm A = 0.0435 \; m^2$

Acceleration due to gravity, $\rm g = 9.81\;m/sec^2$

Solution :

We know that the absolute pressure is given by the formula

$\rm P=P_0+ h \rho g$

a)

$\rm P = P_0 + (h\times1\times 10^3\times 9.8)$

$\rm P = P_0 + 9800h$  ---- (1)

b)

Now, $\rm P_0 = 10^5\;Pa$  and h = 9.5 m

so from equation (1)

$\rm P = 10^5 + (9800\times 9.5)$

P = 193100 Pa

c)

$\rm Pressure = \dfrac{Force }{Area}$

F = PA --- (2)

d)

Now given that, A = 0.0385 $\rm m^2$

and P = 193100 Pa

So from equation (2)

$\rm F = 193100\times 0.0385$

F = 7434.35 N

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