Question #312977

A table tennis ball has a diameter of 3.80 cm and an average density of 0.0840 g/cm3. What force is required to hold it completely submerged underwater?


1
Expert's answer
2022-03-20T18:58:01-0400

Given

dd- Diameter of a tenis ball=3.80cm

ρ\rho- Density of the tenis ball=0.0840g/cm3=0.0840g/cm^3

dwdw- Density of water=1g/cm3=1g/cm^3

Volume of the ball,wv=43πd38w-v=43 \pi d38

=16π×(3.8)3cm3=28.73cm3=16\pi\times(3.8)3cm^3=28.73cm^3

ww- weight of the ball=v×ρ×g=v×\rho ×g

when the ball is submerged the buoyant force acting on the ball wil be equal to the weight of displaced water.

So the bouyant force

Fb=v×dw×gFb=v\times dw \times g

So the net downward force to be appliied on the ball to just submerge it will be

Fnet=Fbw=vdwgvρg=vg(dwρ)F_{net}=Fb-w=vdwg-v\rho g=vg(dw-\rho )

=28.73×980×(10.084)dyne=25790.3dyne=28.73 \times 980 \times (1-0.084) dyne = 25790.3 dyne



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