Since it barely floats, the sphere volume, filled with water, would weigh 100 N (not including the hollow sphere itself).
pi*D^3/6 * 1000 kg/m^3*g 9.8 m/s^2 = 100 N
pi D^3/6 = 100/9800 = 0.010204
D^3 = 0.19488
D = 0.269 m = 26.9 cm
The weight of a hollow sphere is 100N.If it floats in water just fully submerged,what is the exterrnal diameter of the shere ?
2 answers
W = density * g * V
100 = 1000 * 9.81 * (4/3 * pi*r^3)
r = .13900
2r = .269
26.9cm convertion
100 = 1000 * 9.81 * (4/3 * pi*r^3)
r = .13900
2r = .269
26.9cm convertion