Question #153398

A table -tennis ball is held under water and then released. Described and explain the subsequent motion of the ball


Expert's answer

The mass of the ball is 2.7(g)2.7(g)


The diameter of the ball is 40(mm)40 (mm)


Gravity, buoyant force and the resistance force of the medium act on the ball.


FG=mg=0.0027⋅(−9.8)≈−0.03(N)F_G=mg=0.0027\cdot(-9.8)\approx-0.03(N)


FB=ρgV=1000⋅9.8⋅(4/3)⋅π⋅0.023≈0.34(N)F_B=\rho gV=1000\cdot9.8\cdot(4/3)\cdot\pi\cdot0.02^3\approx0.34(N)


FR=−kvF_R=-kv.


So, if the force of gravity and the force of resistance of the medium become equal to the modulus of the buoyant force, the ball will move up at a constant speed.



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