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 : 
5 times that of initial pressure i.e 1625 kpa
Answer:
90
Explanation:
The mass number of the missing daughter nuclei can be obtained as shown in the attached photo.
Answer:
no.
Explanation:
because the mass of an object never changes.
Answer:
It would take approximately 289 hours for the population to double
Explanation:
Recall the expression for the continuous exponential growth of a population:
where N(t) measures the number of individuals, No is the original population, "k" is the percent rate of growth, and "t" is the time elapsed.
In our case, we don't know No (original population, but know that we want it to double in a certain elapsed "t". We also have in mind that the percent rate "k" would be expressed in mathematical form as: 0.0024 (mathematical form of the given percent growth rate).
So we need to solve for "t" in the following equation:
Which can be rounded to about 289 hours
