Originally Posted by **Heavyarms2050** **5. A thin spherical shell of mass 0.400 kg and diameter 0.200 m is filled with alcohol (rho = 806 kg/m3). It is released from rest at the bottom of a tank of water (rho = 1000 kg/m3). What is the acceleration of the alcohol spherical shell as it starts to rise to the surface?**
Ans:1.06 m/s^2
Yea, another clueless one |

I believe you have to use archimedes principle for this question. Do you know what it is?

I will quote it for you.

When a body is wholly or partly immersed in a fluid it experiences and upward force equal to the weight of the fluid displaced. |

Firstly we must calculate the weight of the displaced water which. will be the volume of the sphere multiplied by the density of the water.

$\displaystyle Upthrust = 1000 Kg /m^3 \times \frac{4}{3} \pi (0.1)^3 m^3 \times 9.81 m/s^2 $

now you must draw a free body force diagram

note the mass of the sphere is $\displaystyle 806 Kg /m^3 \times \frac{4}{3} \pi (0.1)^3 m^3 + 0.4 Kg $ this is the mass of the volume of the alcohol plus the mass of the shell

so the net force is $\displaystyle Upthrust - Weight $

using Newton's second law

$\displaystyle Mass \times Acceleration = Upthrust - Weight $

$\displaystyle ma= 1000 Kg /m^3 \times \frac{4}{3} \pi (0.1)^3 m^3 \times 9.81 m/s^2 - m \times 9.81 N$

so $\displaystyle a= \frac{1000 Kg /m^3 \times \frac{4}{3} \pi (0.1)^3 m^3 \times 9.81 m/s^2}{ 806 Kg /m^3 \times \frac{4}{3} \pi (0.1)^3 m^3 + 0.4 Kg } - 9.81 m/s^2$

giving $\displaystyle a= 1.07195 m/s^2$

or $\displaystyle a= 1.06 m/s^2$ quoting to 3 significant figures.

I hope that helps

Bobak