Answer:
The tank is losing [tex]4.976*10^{-4} m^3/s[/tex]
[tex]v_g = 19.81 \ m/s[/tex]
Explanation:
According to the Bernoulli’s equation:
[tex]P_1 + 1 \frac{1}{2} \rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho gh_2[/tex]
We are being informed that both the tank and the hole is being exposed to air :
∴ P₁ = P₂
Also as the tank is voluminous ; we take the initial volume [tex]v_1[/tex] ≅ 0 ;
then [tex]v_2[/tex] can be determined as:[tex]\sqrt{[2g (h_1- h_2)][/tex]
h₁ = 5 + 15 = 20 m;
h₂ = 15 m
[tex]v_2 = \sqrt{[2*9.81*(20 - 15)][/tex]
[tex]v_2 = \sqrt{[2*9.81*(5)][/tex]
[tex]v_2= 9.9 \ m/s[/tex] as it leaves the hole at the base.
radius r = d/2 = 4/2 = 2.0 mm
(a) From the law of continuity; its equation can be expressed as:
J = [tex]A_1v_2[/tex]
J = πr²[tex]v_2[/tex]
J =[tex]\pi *(2*10^{-3})^{2}*9.9[/tex]
J =[tex]1.244*10^{-4} m^3/s[/tex]
b)
How fast is the water from the hole moving just as it reaches the ground?
In order to determine that; we use the relation of the velocity from the equation of motion which says:
v² = u² + 2gh ₂
v² = 9.9² + 2×9.81×15
v² = 392.31
The velocity of how fast the water from the hole is moving just as it reaches the ground is : [tex]v_g = \sqrt{392.31}[/tex]
[tex]v_g = 19.81 \ m/s[/tex]
