Please help me with 1-3
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Answer:
Explanation:
From the given information:
The diameter of the pool = 24 ft
The radius will be = 24 ft/2 = 12 ft
The volume of water V = πr²(Δx)
V = π× 12²×(Δx)
V = 144π(Δx)
Le's assume water weighs 62.5 lb/ft³;
Then:
the Force (F) will be:
= 144π(Δx) * 62.5
= 9000πΔx lb
Also, the side of cylindrical water = 5 ft while its depth = 4ft
As such, each slide of water d = 5 - x, and the region is between 0 and 4.
∴
The required work is:
Here we have a circular motion. In that type of motion two new forces appear: centrifugal and centripetal. Centrifugal force tries to move object away frome the center of rotation. Centripetal force is the reason why object moves in a circle.
According to Newton's first law, in order for water not to spill out total force acting on water must be equal to zero. Forces acting on the water are gravitational and centrifugal.
![0= F_{g} + F_{c} \\ - [tex]F_{g} = F_{c} \\ m*g= \frac{m* v^{2} }{r} \\ g= \frac{ v^{2} }{r} \\ v^{2} =g*r \\ v= \sqrt{g*r} \\ v= \sqrt{9.81*0.95} \\ v= \sqrt{9.3195} \\ v=3.1 m/s](https://tex.z-dn.net/?f=0%3D%20F_%7Bg%7D%20%2B%20F_%7Bc%7D%20%20%5C%5C%20-%20%5Btex%5DF_%7Bg%7D%20%3D%20F_%7Bc%7D%20%5C%5C%20m%2Ag%3D%20%5Cfrac%7Bm%2A%20v%5E%7B2%7D%20%7D%7Br%7D%20%20%5C%5C%20g%3D%20%5Cfrac%7B%20v%5E%7B2%7D%20%7D%7Br%7D%20%5C%5C%20%20v%5E%7B2%7D%20%3Dg%2Ar%20%5C%5C%20%20v%3D%20%5Csqrt%7Bg%2Ar%7D%20%20%5C%5C%20v%3D%20%5Csqrt%7B9.81%2A0.95%7D%20%20%5C%5C%20v%3D%20%5Csqrt%7B9.3195%7D%20%20%5C%5C%20v%3D3.1%20m%2Fs)
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Forces have opposite signs as they act in opposite directions. But their value is the same so we can ignore signs:
According to Newton's first law, in order for water not to spill out total force acting on water must be equal to zero. Forces acting on the water are gravitational and centrifugal.
Forces have opposite signs as they act in opposite directions. But their value is the same so we can ignore signs: