A child playing in a swimming pool realizes that it is easy to push a small inflated ball under the surface of the water whereas a large ball requires a lot of force. The child happens to have a styrofoam ball (the shape of the ball will not distort when it is forced under the surface), which he forces under the surface of the water. If the child needs to supply 608 N to totally submerge the ball, calculate the diameter ???? of the ball. The density of water is ????w=1.000×103 kg/m3, the density of styrofoam is ????foam=95.0 kg/m3, and the acceleration due to gravity is ????=9.81 m/s2.

And the buoyancy is the weight of the fluid displaced (water for us), in our case, this volume of fluid displaced is equal to the volume of the ball:

[tex]B=W_{fd}=m_{fd}g=\rho_{fd}V_{fd}g=\rho_{fd}V_bg[/tex]

Putting all together:

[tex]\rho_{fd}V_bg=\rho_b V_b g + F[/tex]

[tex]\rho_{fd}V_bg-\rho_b V_b g=F[/tex]

[tex]V_bg(\rho_{fd}-\rho_b)=F[/tex]

[tex]V_b=\frac{F}{g(\rho_{fd}-\rho_b)}[/tex]

Using the formula for the volume of a sphere we have then:

[tex]\frac{4\pi r^3}{3}=\frac{F}{g(\rho_{fd}-\rho_b)}[/tex]

[tex]r=\sqrt[3]{\frac{3F}{g4\pi(\rho_{fd}-\rho_b)}}[/tex]

Or in terms of the diameter:

[tex]d=2\sqrt[3]{\frac{3F}{g4\pi(\rho_{fd}-\rho_b)}}[/tex]

Which for our values is:

[tex]d=2\sqrt[3]{\frac{3(608N)}{(9.81m/s^2)4\pi((1\times10^3kg/m^3)-(95kg/m^3))}}=0.5m[/tex]

snehashish65 snehashish65 · Answer 2 · 2021-10-18T14:01:18+02:00

The diameter of ball submerged into water is of 0.5 m.

Given data:

The magnitude of force applied by child is, F = 608 N.

The density of water is, [tex]\rho_{w} = 1000 \;\rm kg/m^{3}[/tex].

The density of styrofoam is, [tex]\rho_{s} = 95 \;\rm kg/m^{3}[/tex].

Gravitational acceleration is, [tex]g = 9.8 \;\rm m/s^{2}[/tex].

As per the concept of buoyancy, the buoyant force (B) applied by the water is in equilibrium with the weight of ball (W) and force applied by the child (F). Therefore,

Buoyant Force = Weight + Force applied

[tex]B = W+F\\\\\rho_{w}gV=\rho_{s}gV+F\\V =\dfrac{F}{(\rho_{w}-\rho_{s})g} \\V =\dfrac{608}{(1000-95) \times 9.8}\\V=0.0685 \;\rm m^{3}[/tex]

The volume of ball is,

[tex]V = \dfrac{4}{3} \pi r^{3}\\[/tex]

Here, r is the radius of ball.

[tex]0.0685= \dfrac{4}{3} \pi \times r^{3}\\r = 0.25 \;\rm m[/tex]

Then diameter of ball is,

[tex]d =2r\\d= 2 \times 0.25\\d =0.5 \;\rm m[/tex]

Thus, the diameter of the ball is 0.5 m.

Learn more about the buoyant force here:

https://brainly.com/question/18103369?referrer=searchResults