Quick answer: 5.4 pounds for a ball 3" in radius.
Remark
You are expected to do this by calling 4/3*pi = some constant k
Givens
r = 2 inches
r1 = 3 inches
Ratio
[tex] \frac{W1}{W2} = \frac{k r^3}{k (r1)^3} [/tex]
Sub and solve
Begin by cancelling out the ks.
[tex] \frac{W1}{W2} = \frac{r^3}{(r1)^3} [/tex]
Next substitute for the 2 rs.
[tex] \frac{W1}{W2} = \frac{2^3}{3^3} [/tex]
Now expand the 2 cubic amounts.
[tex] \frac{W1}{W2} = \frac{8}{27} [/tex]
Conclusion Part One
The ratio of the two weights is 8/27
Part Two
Set up the ratio for 1.6 pounds to x
[tex] \frac{8}{27} = \frac{1.6}{x} [/tex]
Cross multiply
8*x = 27 * 1.6 Combine the right side.
8x = 43.2 Divide by 8
x = 43.2 / 8
x = 5.4 pounds.
If you take physics, you will do this quite frequently.
