To solve this problem it is necessary to apply the concepts related to the geometry of a cylindrical tank and its respective definition.
The volume of a tank is given by
![V = \frac{\pi d^2}{4}h](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7B%5Cpi%20d%5E2%7D%7B4%7Dh)
Where
d = Diameter
h = Height
Considering that there are two stages, let's define the initial and final volume as,
![V_0 = \frac{\pi d^2}{4}H](https://tex.z-dn.net/?f=V_0%20%3D%20%5Cfrac%7B%5Cpi%20d%5E2%7D%7B4%7DH)
![V_f = \frac{\pi d^2}{4}h](https://tex.z-dn.net/?f=V_f%20%3D%20%5Cfrac%7B%5Cpi%20d%5E2%7D%7B4%7Dh)
We know as well by definition that
![1gal = 3.785*10^{-3}m^3](https://tex.z-dn.net/?f=1gal%20%3D%203.785%2A10%5E%7B-3%7Dm%5E3)
Then we have for the statement that
![V_f = V_0 -1gal](https://tex.z-dn.net/?f=V_f%20%3D%20V_0%20-1gal)
![V_f = V_0 - 3.785*10^{-3}](https://tex.z-dn.net/?f=V_f%20%3D%20V_0%20-%203.785%2A10%5E%7B-3%7D)
Replacing the previous data
![\frac{\pi d^2}{4}h = \frac{\pi d^2}{4}H- 3.785*10^{-3}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cpi%20d%5E2%7D%7B4%7Dh%20%3D%20%5Cfrac%7B%5Cpi%20d%5E2%7D%7B4%7DH-%203.785%2A10%5E%7B-3%7D)
![\frac{\pi (3.6)^2}{4}h = \frac{\pi (3.6)^2}{4}(2)- 3.785*10^{-3}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cpi%20%283.6%29%5E2%7D%7B4%7Dh%20%3D%20%5Cfrac%7B%5Cpi%20%283.6%29%5E2%7D%7B4%7D%282%29-%203.785%2A10%5E%7B-3%7D)
Solving to get h,
![h = 1.99963m](https://tex.z-dn.net/?f=h%20%3D%201.99963m)
Therefore the change is
![\Delta h = H-h](https://tex.z-dn.net/?f=%5CDelta%20h%20%3D%20H-h)
![\Delta h = 2- 1.99963](https://tex.z-dn.net/?f=%5CDelta%20h%20%3D%202-%201.99963)
![\Delta h = 3.7*10^{-4}m=0.37mm](https://tex.z-dn.net/?f=%5CDelta%20h%20%3D%203.7%2A10%5E%7B-4%7Dm%3D0.37mm)
Therefore te change in the height of the water in the tank is 0.37mm
Answer:
The velocity of the particle = -1.92 m/s
The speed of the particle = 5.72 m/s
Explanation:
Given equation of motion;
![f(t) = 18 \ + \ \frac{48}{t} \ + \ 1](https://tex.z-dn.net/?f=f%28t%29%20%3D%2018%20%5C%20%2B%20%5C%20%5Cfrac%7B48%7D%7Bt%7D%20%5C%20%2B%20%5C%201)
Velocity is defined as the change in displacement with time.
![V = \frac{df(t)}{dt} = -\frac{48}{t^2} \\\\at \ t = 5 \ s\\\\V = -\frac{48}{5^2} = \frac{-48}{25} = - 1.92 \ m/s](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7Bdf%28t%29%7D%7Bdt%7D%20%3D%20-%5Cfrac%7B48%7D%7Bt%5E2%7D%20%5C%5C%5C%5Cat%20%5C%20t%20%3D%205%20%5C%20s%5C%5C%5C%5CV%20%3D%20-%5Cfrac%7B48%7D%7B5%5E2%7D%20%3D%20%5Cfrac%7B-48%7D%7B25%7D%20%3D%20-%201.92%20%5C%20m%2Fs)
The distance traveled by the particle in 5 s:
![s = f(5) = 18 + \frac{48}{5} + 1\\\\s= 28.6 \ m](https://tex.z-dn.net/?f=s%20%3D%20f%285%29%20%3D%2018%20%2B%20%5Cfrac%7B48%7D%7B5%7D%20%2B%201%5C%5C%5C%5Cs%3D%2028.6%20%5C%20m)
The speed of the particle when t = 5s
![Speed = \frac{28.6}{5} = 5.72 \ m/s](https://tex.z-dn.net/?f=Speed%20%3D%20%5Cfrac%7B28.6%7D%7B5%7D%20%3D%205.72%20%5C%20m%2Fs)
25% i believe because if were talking 50 percent half it would be 25.
Answer:
answered it.................