Answer:
The pressure is 
Explanation:
From the question we are told that
The gauge pressure at the mouth is 
The radius of the column is 
The speed of the liquid outside the body is 
The area of the column is 
The area inside the mouth 
Generally according to continuity equation
=> 
=> 
=> 
So
=> 
=> 
substituting values
Now the height of inside the mouth is 
Now the height of the column is 
Generally according to Bernoulli's equation
![p_1 = [\frac{1}{2} \rho v_2^2 + h_2 \rho g +p_2] -[\frac{1}{2} \rho * v_1^2 + h_1 \rho g ]](https://tex.z-dn.net/?f=p_1%20%3D%20%20%5B%5Cfrac%7B1%7D%7B2%7D%20%20%5Crho%20v_2%5E2%20%2B%20h_2%20%5Crho%20g%20%2Bp_2%5D%20-%5B%5Cfrac%7B1%7D%7B2%7D%20%5Crho%20%2A%20v_1%5E2%20%2B%20h_1%20%5Crho%20g%20%5D)
Now 
which is the density of water
is the gauge pressure of the atmosphere which is zero
So
![p_1 = [(0.5 * 1000 * (3.1)^2) +(0.008 * 1000 * 9.8) + 0]-](https://tex.z-dn.net/?f=p_1%20%3D%20%20%5B%280.5%20%2A%201000%20%2A%20%283.1%29%5E2%29%20%2B%280.008%20%2A%201000%20%2A%209.8%29%20%2B%200%5D-)
![[(0.5 * 1000 * 0.31^2) + (0.08*1000 * 9.8)]](https://tex.z-dn.net/?f=%5B%280.5%20%2A%201000%20%2A%200.31%5E2%29%20%2B%20%280.08%2A1000%20%2A%209.8%29%5D)
I'm going to assume this is over a horizontal distance. You know from Newton's Laws that F=ma --> a = F/m. You also know from your equations of linear motion that v^2=v0^2+2ad. Combining these two equations gives you v^2=v0^2+2(F/m)d. We can plug in the given values to get v^2=0^2+2(20/3)0.25. Solving for v we get v=1.82 m/s!
Answer:
The water level rises more when the cube is located above the raft before submerging.
Explanation:
These kinds of problems are based on the principle of Archimedes, who says that by immersing a body in a volume of water, the initial water level will be increased, raising the water level. That is, the height in the container with water will rise in level. The difference between the new volume and the initial volume of the water will be the volume of the submerged body.
Now we have two moments when the steel cube is held by the raft and when it is at the bottom of the pool.
When the cube is at the bottom of the water we know that the volume will increase, and we can calculate this volume using the volume of the cube.
Vc = 0.45*0.45*0.45 = 0.0911 [m^3]
Now when a body floats it is because a balance is established in the densities, the density of the body and the density of the water.
![Ro_{H2O}=R_{c+r}\\where:\\Ro_{H2O}= water density = 1000 [kg/m^3]\\Ro_{c+r}= combined density cube + raft [kg/m^3]](https://tex.z-dn.net/?f=Ro_%7BH2O%7D%3DR_%7Bc%2Br%7D%5C%5Cwhere%3A%5C%5CRo_%7BH2O%7D%3D%20water%20density%20%3D%201000%20%5Bkg%2Fm%5E3%5D%5C%5CRo_%7Bc%2Br%7D%3D%20combined%20density%20cube%20%2B%20raft%20%5Bkg%2Fm%5E3%5D)
Density is given by:
Ro = m/V
where:
m= mass [kg]
V = volume [m^3]
The buoyancy force can be calculated using the following equation:
![F_{B}=W=Ro_{H20}*g*Vs\\W = (200+730)*9.81\\W=9123.3[N]\\\\9123=1000*9.81*Vs\\Vs = 0.93 [m^3]](https://tex.z-dn.net/?f=F_%7BB%7D%3DW%3DRo_%7BH20%7D%2Ag%2AVs%5C%5CW%20%3D%20%28200%2B730%29%2A9.81%5C%5CW%3D9123.3%5BN%5D%5C%5C%5C%5C9123%3D1000%2A9.81%2AVs%5C%5CVs%20%3D%200.93%20%5Bm%5E3%5D)
Vs > Vc, What it means is that the combined volume of the raft and the cube is greater than that of the cube at the bottom of the pool. Therefore the water level rises more when the cube is located above the raft before submerging.
