### Video Transcript

Water flows smoothly through a pipe at a speed of 1.66 meters per second. The water then flows smoothly from the first pipe into and through a second pipe at a speed of 1.22 meters per second. What is the ratio of the cross-sectional area of the first pipe to that of the second pipe?

Weโll start by drawing a diagram. We have two pipes. The first pipe has a cross-sectional area that we will call ๐ด one, and the second pipe has a cross-sectional area that we will call ๐ด two. These pipes are connected so that any fluid that passes through the first pipe also passes through the second pipe. In this case, we are told that water flows through the pipes, flowing through the first pipe with a speed that we will call ๐ฃ one and through the second pipe with a speed that we will call ๐ฃ two.

The question asked us to calculate the ratio of the cross-sectional area of the first pipe to that of the second pipe. So we must calculate ๐ด one divided by ๐ด two. The question tells us that water flows through the first pipe with a speed of 1.66 meters per second. So ๐ฃ one is equal to 1.66 meters per second. The question goes on to tell us that the water flows through the second pipe with a speed of 1.22 meters per second. So ๐ฃ two is equal to 1.22 meters per second. And these are both in SI units, so we donโt need to worry about converting any of them before substituting them into any equations.

We will answer this question using the continuity equation for fluids, which states that a fluidโs density multiplied by the cross-sectional area of the pipe it is flowing in multiplied by the fluidโs speed is constant. This means that we can write the waterโs density multiplied by the cross-sectional area of the pipe it is flowing in multiplied by the waterโs speed in the first pipe is equal to the waterโs density multiplied by the cross-sectional area of the pipe it is flowing in multiplied by the speed of the water in the second pipe.

Because the fluid is water, we can say that it is incompressible, meaning that its density is constant. So the density of the water in the first pipe is equal to the density of the water in the second pipe. And we will call this overall density ๐. And we can simplify our equation by dividing both sides by ๐, where we see that the ๐โs on the left and on the right cancel, leaving us with ๐ด one multiplied by ๐ฃ one is equal to ๐ด two multiplied by ๐ฃ two.

And we would like to rearrange this to get an expression for ๐ด one divided by ๐ด two. We will start by dividing both sides of the equation by ๐ฃ one, where we see that the ๐ฃ ones on the left cancel. Next, we will divide both sides of the equation by ๐ด two, where we see that the ๐ด twos on the right cancel. And this gives us our expression of the ratio of the cross-sectional area of the first pipe to that of the second pipe. ๐ด one divided by ๐ด two is equal to ๐ฃ two divided by ๐ฃ one.

We can now substitute unknown values of ๐ฃ one and ๐ฃ two into this equation. ๐ด one divided by ๐ด two is equal to 1.22 meters per second divided by 1.66 meters per second. Evaluating this expression gives ๐ด one divided by ๐ด two is equal to 0.735 to three decimal places. The ratio of the cross-sectional area of the first pipe to that of the second pipe is equal to 0.735 to three decimal places.