Answer:
A
Explanation:
Becuse its complete number
Answer:
The depth is 5.15 m.
Explanation:
Lets take the depth of the pool = h m
The atmospheric pressure ,P = 101235 N/m²
The area of the top = A m²
The area of the bottom = a m²
Given that A= 1.5 a
The force on the top of the pool = P A
The total pressure on the bottom = P + ρ g h
ρ =Density of the water = 1000 kg/m³
The total pressure at the bottom of the pool = (P + ρ g h) a
The bottom and the top force is same
(P + ρ g h) a = P A
P a +ρ g h a = P A
ρ g h a = P A - P a
![h=\dfrac{P ( A-a)}{\rho g a}](https://tex.z-dn.net/?f=h%3D%5Cdfrac%7BP%20%28%20A-a%29%7D%7B%5Crho%20g%20a%7D)
![h=\dfrac{P ( 1.5 a-a)}{\rho g a}](https://tex.z-dn.net/?f=h%3D%5Cdfrac%7BP%20%28%201.5%20a-a%29%7D%7B%5Crho%20g%20a%7D)
![h=\dfrac{P ( 1.5- 1)}{\rho g}](https://tex.z-dn.net/?f=h%3D%5Cdfrac%7BP%20%28%201.5-%201%29%7D%7B%5Crho%20g%7D)
![h=\dfrac{101235 ( 1.5- 1)}{1000\times 9.81}\ m](https://tex.z-dn.net/?f=h%3D%5Cdfrac%7B101235%20%28%201.5-%201%29%7D%7B1000%5Ctimes%209.81%7D%5C%20m)
h=5.15 m
The depth is 5.15 m.
The answer & explanation for this question is given in the attachment below.
To solve this exercise we need the concept of Kinetic Energy and its respective change: Initial and final kinetic energy.
Let's start considering that the angular velocity is given by,
![\omega = \frac{v}{R}](https://tex.z-dn.net/?f=%5Comega%20%3D%20%5Cfrac%7Bv%7D%7BR%7D)
Where,
V = linear speed
R = the radius
In the case of the initial kinetic energy:
![KE_i=\frac{1}{2} mv^2 + \frac{1}{2}I \omega^2](https://tex.z-dn.net/?f=KE_i%3D%5Cfrac%7B1%7D%7B2%7D%20mv%5E2%20%2B%20%5Cfrac%7B1%7D%7B2%7DI%20%5Comega%5E2)
Where I is the moment of inertia previously defined.
![KE_i = \frac{1}{2}(m)3.5^2 + \frac{1}{2}* (\frac{2}{5} m R^2) (\frac{3.5}{R})^2](https://tex.z-dn.net/?f=KE_i%20%3D%20%5Cfrac%7B1%7D%7B2%7D%28m%293.5%5E2%20%2B%20%5Cfrac%7B1%7D%7B2%7D%2A%20%28%5Cfrac%7B2%7D%7B5%7D%20m%20R%5E2%29%20%28%5Cfrac%7B3.5%7D%7BR%7D%29%5E2)
In the case of the final kinetic energy, we have to,
![KE_f= mgh+ \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2](https://tex.z-dn.net/?f=KE_f%3D%20mgh%2B%20%5Cfrac%7B1%7D%7B2%7D%20mv%5E2%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20I%20%5Comega%5E2)
![KE_f = m * 9.81 * 0.76 + \frac{1}{2} m v^2 + \frac{1}{2} (\frac{2}{5} m R^2) (\frac{v}{R})^2](https://tex.z-dn.net/?f=KE_f%20%3D%20m%20%2A%209.81%20%2A%200.76%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20m%20v%5E2%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20%28%5Cfrac%7B2%7D%7B5%7D%20m%20R%5E2%29%20%28%5Cfrac%7Bv%7D%7BR%7D%29%5E2)
For conservation of Energy we have, that
, then (canceling the mass and the radius)
![\frac{1}{2} 3.5^2 + \frac{1}{2}(\frac{2}{5})(3.5)^2= 9.81 * 0.76 + \frac{1}{2} v^2 + \frac{1}{2} (\frac{2}{5}) (v)^2](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%203.5%5E2%20%2B%20%5Cfrac%7B1%7D%7B2%7D%28%5Cfrac%7B2%7D%7B5%7D%29%283.5%29%5E2%3D%209.81%20%2A%200.76%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20v%5E2%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20%28%5Cfrac%7B2%7D%7B5%7D%29%20%28v%29%5E2)
![8.575= 7.4556+ \frac{1}{2} v^2 + \frac{1}{2} (\frac{2}{5}) (v)^2](https://tex.z-dn.net/?f=8.575%3D%207.4556%2B%20%5Cfrac%7B1%7D%7B2%7D%20v%5E2%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20%28%5Cfrac%7B2%7D%7B5%7D%29%20%28v%29%5E2)
![1.1194= \frac{1}{2}( v^2 + (\frac{2}{5}) (v)^2)](https://tex.z-dn.net/?f=1.1194%3D%20%5Cfrac%7B1%7D%7B2%7D%28%20v%5E2%20%2B%20%28%5Cfrac%7B2%7D%7B5%7D%29%20%28v%29%5E2%29)
![2.2388= (\frac{7}{5}) (v)^2](https://tex.z-dn.net/?f=2.2388%3D%20%28%5Cfrac%7B7%7D%7B5%7D%29%20%28v%29%5E2)
![v=1.26m/s](https://tex.z-dn.net/?f=v%3D1.26m%2Fs)
The quantities which are measured is called physical quantity.
Ex - Time, Length, Density. etc.
I hope it's help you..
Thanks♥♥