Question

The fish tank has side lengths 20in, 10in and height 15in. The water level is two inches below the top of the tank. A glass sphere of radius 1in is dropped in to the tank. What is the new distance from the water to the top of the tank? How many of these balls can be put into the tank with the tank not overflowing?

Answer 1

Answer:

Part 1) The new distance from the water to the top of the tank is $1.979\ in$

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

$V=20*10*15=3,000\ in^{3}$

step 2

Find the volume of the tank if the water level is two inches below the top of the tank

$V=20*10*(15-2)=2,600\ in^{3}$

step 3

Find the volume of the glass sphere

The volume of the glass sphere is equal to

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

we have

$r=1\ in$

assume

$\pi=3.14$

substitute

$V=\frac{4}{3}(3.14)(1)^{3}$

$V=4.2\ in^{3}$

step 4

What is the new distance from the water to the top of the tank?

we know that

$2 in -----> represent (3,000-2,600)=400\ in^{3}$

so

using proportion

Find how many inches correspond a volume of $4.2\ in^{3}$

$\frac{2}{400}\frac{in}{in^{3}}=\frac{x}{4.2}\frac{in}{in^{3}}\\ \\x=4.2*2/400\\ \\x=0.021\ in$

The new distance from the water to the top of the tank is

$2-0.021=1.979\ in$

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  $4.2\ in^{3}$

using proportion

$\frac{1}{4.2}=\frac{x}{400}\\ \\x=400/4.2\\ \\x=95.23\ balls$

therefore

The maximum number of balls that can be put into the tank with the tank not overflowing is 95