4.7 cm. The tube is sealed at one end and loaded with lead shot to give it a total mass of M = 130 g. The tube floats in water (of density 1 g/cm3 ) in vertical position, loaded end down. What is the depth of the bottom end of the tube? Answer in units of cm.

Depth is 7.49cm

Explanation:

Given that;

The diameter of tube is (D)=4.7cm

Area of the tube is =πD^2/4

Area is 3.142*4.7*4.7/4=17.35cm2

Mass(M)=130g

Density =1g/cm^3

Recall that Density =mass /volume

Hence volume =130/1=130cm^3

Also note

Volume=area*depth

130=17.35*depth

Depth=130/17.35=7.49cm

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft