Answer:
Part a)

Part b)

Part c)
Since we know that the base area will remain same always
so here the length and width of the object is not necessary to obtain the above data in such type of questions
Explanation:
Part a)
As we know that when cylinder float in the water then weight of the cylinder is counter balanced by the buoyancy force
So here we know
buoyancy force is given as



Now we know that the weight of the cylinder is given as

now we have


Part b)
When the same cylinder is floating in other liquid then we will have

so we have


Part c)
Since we know that the base area will remain same always
so here the length and width of the object is not necessary to obtain the above data in such type of questions
Answer:_COC1\/2+_H\/2O>_HC1+CO\/2
Explanation:
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Answer:

Explanation:
The difference between top and bottom surfaces is computed by the following hydrostatic equivalence:



Answer:
5488 Joules
Explanation:
Use the formula for the potential energy:

Answer:
the height reached is = 0.458 [m]
Explanation:
We need to make a sketch of the ball and see the location of the reference point where the potential energy is zero. But the kinetic energy will be defined by the following expression:
![Ek=\frac{1}{2} *m*v^{2} \\where:Ek= kinetic energy [J]\\m = mass of the ball [kg]\\v = velocity of the ball [m/s]](https://tex.z-dn.net/?f=Ek%3D%5Cfrac%7B1%7D%7B2%7D%20%2Am%2Av%5E%7B2%7D%20%5C%5Cwhere%3AEk%3D%20kinetic%20energy%20%5BJ%5D%5C%5Cm%20%3D%20mass%20of%20the%20ball%20%5Bkg%5D%5C%5Cv%20%3D%20velocity%20of%20the%20ball%20%5Bm%2Fs%5D)
Replacing the values on the equation we have:
![Ek=\frac{1}{2}*(2)*(3^{2} )\\ Ek=9[J]\\](https://tex.z-dn.net/?f=Ek%3D%5Cfrac%7B1%7D%7B2%7D%2A%282%29%2A%283%5E%7B2%7D%20%29%5C%5C%20Ek%3D9%5BJ%5D%5C%5C)
This kinetic energy will be transformed in potential energy in the moment when the ball starts to rolling up. Therefore the maximum height reached by the ball depends of the initial velocity given to the ball.
![Ek=Ep\\where\\Ep=potential energy [J]\\Ep=m*g*h\\where\\g=gravity = 9.81[m/s^2]\\h=height reached [m]\\](https://tex.z-dn.net/?f=Ek%3DEp%5C%5Cwhere%5C%5CEp%3Dpotential%20energy%20%5BJ%5D%5C%5CEp%3Dm%2Ag%2Ah%5C%5Cwhere%5C%5Cg%3Dgravity%20%3D%209.81%5Bm%2Fs%5E2%5D%5C%5Ch%3Dheight%20reached%20%5Bm%5D%5C%5C)
Now we have:
![h=\frac{Ep}{m*g} \\h=\frac{9}{2*9.81} \\\\h=0.45 [m]](https://tex.z-dn.net/?f=h%3D%5Cfrac%7BEp%7D%7Bm%2Ag%7D%20%5C%5Ch%3D%5Cfrac%7B9%7D%7B2%2A9.81%7D%20%5C%5C%5C%5Ch%3D0.45%20%5Bm%5D)
In that moment when the ball reach the 0.45 [m] the potencial energy will be maximum and equal to the kinetic energy when the ball has a velocity of 3[m/s]