Answer:
Part 1) The new distance from the water to the top of the tank is [tex]1.979\ in[/tex]
Part 2) The maximum number of balls that can be put into the tank with the tank not overflowing is 95
Step-by-step explanation:
step 1
Find the total volume of the tank
[tex]V=20*10*15=3,000\ in^{3}[/tex]
step 2
Find the volume of the tank if the water level is two inches below the top of the tank
[tex]V=20*10*(15-2)=2,600\ in^{3}[/tex]
step 3
Find the volume of the glass sphere
The volume of the glass sphere is equal to
[tex]V=\frac{4}{3}\pi r^{3}[/tex]
we have
[tex]r=1\ in[/tex]
assume
[tex]\pi=3.14[/tex]
substitute
[tex]V=\frac{4}{3}(3.14)(1)^{3}[/tex]
[tex]V=4.2\ in^{3}[/tex]
step 4
What is the new distance from the water to the top of the tank?
we know that
[tex]2 in -----> represent (3,000-2,600)=400\ in^{3}[/tex]
so
using proportion
Find how many inches correspond a volume of [tex]4.2\ in^{3}[/tex]
[tex]\frac{2}{400}\frac{in}{in^{3}}=\frac{x}{4.2}\frac{in}{in^{3}}\\ \\x=4.2*2/400\\ \\x=0.021\ in[/tex]
The new distance from the water to the top of the tank is
[tex]2-0.021=1.979\ in[/tex]
step 5
Find how many of these balls can be put into the tank with the tank not overflowing
we know that
The volume of one ball is equal to [tex]4.2\ in^{3}[/tex]
using proportion
[tex]\frac{1}{4.2}=\frac{x}{400}\\ \\x=400/4.2\\ \\x=95.23\ balls[/tex]
therefore
The maximum number of balls that can be put into the tank with the tank not overflowing is 95
