Answer:
The tank is losing

Explanation:
According to the Bernoulli’s equation:
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
≅ 0 ;
then
can be determined as:![\sqrt{[2g (h_1- h_2)]](https://tex.z-dn.net/?f=%5Csqrt%7B%5B2g%20%28h_1-%20h_2%29%5D)
h₁ = 5 + 15 = 20 m;
h₂ = 15 m
![v_2 = \sqrt{[2*9.81*(20 - 15)]](https://tex.z-dn.net/?f=v_2%20%3D%20%5Csqrt%7B%5B2%2A9.81%2A%2820%20-%2015%29%5D)
![v_2 = \sqrt{[2*9.81*(5)]](https://tex.z-dn.net/?f=v_2%20%3D%20%5Csqrt%7B%5B2%2A9.81%2A%285%29%5D)
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 = 
J = πr²
J =
J =
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 : 

Answer:
a)
Weight in Air = 0.3N
Weight in Water = 0.25N
Weight in Liquid = 0.24N.
Upthrust /Buoyant Force = Weight in Air – Weight in Fluid(Water in this case)
= 0.3 – 0.25
= 0.5N.
b) R.D of Body = Density of Body/Density of Standard Fluid(Water).
There's a Derived Formula for RD.
I'm gonna Apply it here.
Ask me for the derivation in the Comment section if you need it.
RD = α/ρ = (Weight in Air) / (Upthrust Force)
Where
α = density of the Body(or reference substance)
ρ = density of standard fluid (water)
= 0.3/0.05 = 6.
c) RD of Liquid = (Density of Liquid) /(Density of standard Fluid(water)
Or we just go by that formula
RD of Liquid = Weight in Air/Upthrust(In Liquid)
We'll be using the Upthrust in that Liquid now.
= 0.3 – 0.24 = 0.06
RD = 0.3/0.06 = 5.
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