Question

please help!!!

A child playing in a swimming pool realizes that it is easy to pash a small inflated ball under the surface of the water whereas large ball requires a lot of force. The child happens to have a styrofoam ball (he 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 581 N to totally submerge the ball,calculate the diameter d of the ball. The density of water is pw 1.000 x 10kg/m, the density of styrofoam is Pfos 95.0 kgm3 and the acceleration due to gravity is g -981 m/2 2.14 x1011 ms

Answer 1

When the ball is fully submerged under water , its weight and applied force is acting downwards

the buoyant force is acting upwards

Net force =zero

Weight of ball + Applied force =Buoyant force

$W_{ball}+F=F_b$

$m_{ball}*g+F=\rho _{water}*V_{displaced}*g$

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Density =Mass/Volume

Mass =Density *volume

and displaced volume is same as volume of ball

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$\rho_{ball}*V_{ball}*g+F=\rho _{water}*V_{ball}*g$

95kg/m* Vball *9.81ms 581N 1000kg/m3 * Vball 9.81m/s

95 * Vialj 9.81 581 - 1000* Vall*9.81

931.95Vball 5819810Vball

$581=9810V_{ball}-931.95V_{ball}$

581 8878.051 ball

581 Vball $78.5 8878.05

581 8878,05

3581 4π * 8878.05

0.245772

Diameter,

d2r2 0.245m

ANSWER:

d 0.4977,

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