Question

4.

Answer 1

Answer:

$574.62\ in^3$

Step-by-step explanation:

First we calculate the volume of the cylinder.

$V=\pi r^2*l$

Where r is the radius and l is the length of the cylinder.

We know that:

$r = \frac{diameter}{2}$

$r = \frac{4}{2}$

$r = 2\ in$

Then:

$V=3.14* 2^2*48$

$V=602.88\ in^3$

Assuming that the cannon balls are spherical then the volume of the 2 spheres is:

$V=2*\frac{4}{3}\pi r^3$

$V=2*\frac{4}{3}(3.14)(\frac{3}{2})^3$

$V=2*\frac{4}{3}(3.14)(\frac{3}{2})^3$

$V=28.26\ in^3$

So the space left inside the cannon is

$V=602.88\ in^3 - 28.26\ in^3\\\\V=574.62\ in^3$