Agraduated cylinder is filled to the 40.00-ml mark with a mineral oil. the masses of the cylinder before and after the addition of the mineral oil are 124.966 g and 159.446 g, respectively. in a separate experiment, a metal ball bearing of mass 18.713 g is placed in the cylinder and the cylinder is again filled to the 40.00-ml mark with the mineral oil. the combined mass of the ball bearing and mineral oil is 50.952 g. calculate the density and radius of the ball bearing (volume of a sphere of radius r is 4/3ïr^3

Firstly, Let's experiment this !

Experiment 1 :

159.446g - 124.966g = 34.48g

34.48g = The mass of Mineral oil.

The density of the mineral oil = M/V = 34.48g/40mL = 0.862g/cm³.

Experiment 2 :

124.966 + 18.173 = 143.139 = The mass of solid + cylinder.

124.966 + 50.952 = 175.918 = The mass of solid + cylinder + Mineral water.

175.918 - 143.139 = 32.779 = The mass of added mineral oil.

Explanation:

Now we have to find the volume of the added mineral oil using the density from experiment 1.

V = 32.779g/0.862g/cm³ = 38.02668213mL

Since we found the volume of the solid, we then have to subtract the added mineral oil volume from the total volume from experiment 1.

Volume of solid = 40-38.02668213 = 1.97331787mL

Density of solid = 18.713g/1.97331787mL = 9.483013499g/cm^3

1.97331787 = (4/3)(3.14)r³

1.97331787*(3/4)(3.14) = .4713338861

.4713338861 = r ³

r = 0.7782328425158433

r = 0.78

Now that's our final answer ! r = 0.78

a is the answer to this question

explanation: