Complete Question
The complete question is shown on the first uploaded image
Answer:
The gauge pressure that water has at the House A [tex]P_A = 257020.68 Pa[/tex]
The gauge pressure that water has at the House B [tex]P_B = 188454 \ Pa[/tex]
Explanation:
From the question we are told that
The mass of water when full is [tex]m_f = 7.41* 10^{5} kg[/tex]
Generally the volume of water in this tank is mathematically represented as
[tex]V = \frac{m }{\rho}[/tex]
Where [tex]\rho[/tex] is the density of water with a value of with a value of [tex]\rho = 1000 kg /m^3[/tex]
substituting values
[tex]V = \frac{7.41 *10^5}{10^3}[/tex]
[tex]V = 741 m^3[/tex]
This volume is the volume of a sphere since the tank is spherical so
[tex]V = \frac{4 \pi ^3}{3}[/tex]
making r the subject of the formula
[tex]r =\sqrt[3]{ \frac{741 *3 }{4\pi} }[/tex]
[tex]r = 5.6134 m[/tex]
Now we can use this parameter to obtain the diameter
So
[tex]d = 2 * r[/tex]
substituting values
[tex]d = 2 * 5.6134[/tex]
[tex]d = 11.23m[/tex]
The pressure the water has at faucet in House A is mathematically evaluated as
[tex]P_A = \rho g h_A[/tex]
This height is obtained as follows
[tex]h_A = d+ 15[/tex]
The value 15 is gotten from the diagram
so
[tex]h_A = 15 + 11.23[/tex]
[tex]h_A = 26.22 m[/tex]
Now substituting values
[tex]P_A = 26.23 * 9.8 * 1000[/tex]
[tex]P_A = 257020.68 Pa[/tex]
The pressure the water has at faucet in House B is mathematically evaluated as
[tex]P_B = \rho g h_B[/tex]
This height is obtained as follows
[tex]h_B = d+ 15[/tex]
The value 15 is gotten from the diagram
so
[tex]h_B = d + 15 -h[/tex]
substituting values
[tex]h_B =11.23 + 15 -7[/tex]
[tex]h_A = 19.23 m[/tex]
Now substituting values
[tex]P_B = 19.23 * 9.8 * 1000[/tex]
[tex]P_B = 188454 \ Pa[/tex]
