Answer:
Explanation:
Answer:
0.8712 N
Explanation:
diameter = 5.7 cm
radius, r = 2.85 cm
Volume of sphere, V = 4/3 x 3.14 x 2.85 x 2.85 x 2.85 x 10^-6 = 9.7 x 10^-5 m^3
Density of ball, d = 0.0835 g/cm^3 = 83.5 kg/m^3
density of water, D = 1000 kg/m^3
True weight, W = Volume x density x g = 9.7 x 10^-5 x 83.5 x 9.8
W = 0.07938 N
Buoyant force, B = Volume x density of water x g
B = 9.7 x 10^-5 x 1000 x 9.8 = 0.9506 N
Force required to hold it
F = W - B
F = 0.9506 - 0.07938
F = 0.8712 N
Thus, the force required to hold it is 0.8712 N.
Answer:
0.488 N
Explanation:
Given , The ball has diameter (d)=5.70 cm = 0.0570 m
radius(r) = d/2 = 2.85 cm = 0.0285 m
Average density[tex]=0.0835g/cm^3 = 83.5 kg/m^3[/tex]
There are two forces acting on the ball.. i.e. The buoyancy force and weight of the ball.
The force required to completely submerged under water= Buoyancy force - Weight of ball
buoyancy force = volume×density of water×g
[tex]=\frac{4}{3} \pi r^{3}dg[/tex]
[tex]=4.18\times(0.0285)^3\times1000\times9.8[/tex] [ density of water = 1000 kg/m3]
=0.532 N
mass of ball = average density×Volume
[tex]=83.5\times4.18\times(0.0285)^3[/tex]
= 0.004529 kg
Weight of ball = mg = 0.004529×9.81 =0.044 N
Required force = 0.532 - 0.044 = 0.488 N.
