# Question Video: Calculating a Liquid’s Velocity from Mass Flow Rate Physics • 9th Grade

A fire hose has a length of 15 m and a cross-sectional area of 2.4 × 10⁻⁴ m². The hose ejects 1.6 kg of water that has a density of 1,025 kg/m³ per second. What is the time interval between the water entering one end of the hose and leaving the opposite end?

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### Video Transcript

A fire hose has a length of 15 meters and a cross-sectional area of 2.4 times 10 to the minus four meters squared. The hose ejects 1.6 kilograms of water that has a density of 1,025 kilograms per meter cubed per second. What is the time interval between the water entering one end of the hose and leaving the opposite end?

In this question, we have a hose that has a length of 15 meters, which we will denote with a capital 𝐿. The hose has a cross-sectional area of 2.4 times 10 to the minus four meters squared, and we’ll denote this with a capital 𝐴. Water is flowing through the hose, and we are told that each second 1.6 kilograms of water exits the hose. So, if we were to place some sort of container, such as a bucket, at the end of the hose and then wait for one second, we would collect 1.6 kilograms of water in this bucket. And we will denote this mass with an 𝑚. Let’s also make a note of the time that we’ve collected this water fall with a capital 𝑇. The final piece of information that the question gives us is the density of the water, which is 1,025 kilograms per meter cubed. And we’ll denote this with a 𝜌.

The question would like us to work out what the time interval is between the water entering one end of the hose and leaving the opposite end. If we call this time interval 𝑇 sub ℎ, our objective is to calculate 𝑇 sub ℎ. Now, we can do this if we consider the speed of the water, which we will call 𝑉. We know that the time that an object takes to travel a certain distance is equal to that distance divided by the speed at which it travels. Or, in our case, the time interval, 𝑇 sub ℎ, is equal to the length of the hose, 𝐿, divided by the speed that the water flows, 𝑉. Therefore, in order to work out this time interval, first we need to calculate the speed of the water. And for now, we’ll keep a note of our equation for 𝑇 sub ℎ over here on the right.

We can work out the speed of the water by considering its mass flow. The mass flow of a liquid is the mass of liquid that flows through some cross section per unit of time. And recall that this is equal to the density of the liquid multiplied by the area of the cross section it is flowing through multiplied by the speed of the liquid. In our case, we already have values for 𝑚, 𝑇, 𝜌, and 𝐴. So, if we want to find the speed of the water, all we have to do is rearrange this equation to make 𝑉 the subject. To do this, we’ll take our equation for mass flow and divide both sides by the density of the water 𝜌. And we see that these 𝜌’s on the right cancel.

Next, we’ll divide both sides by the cross-sectional area of the hose 𝐴. And we see that our 𝐴s on the right cancel as well. And this gives us our equation for the speed of the water. Let’s write this a little bit more neatly. So the speed of the water, 𝑉, is equal to the mass of water, 𝑚, that flowed through the hose over some amount of time divided by the density of the water, 𝜌, multiplied by the cross-sectional area of the hose, capital 𝐴, multiplied by the amount of time we observed the water fall, capital 𝑇. Now, we can go ahead and substitute our values for 𝑚, 𝜌, 𝐴, and 𝑇 into this equation. And this gives us 𝑉 is equal to 1.6 kilograms divided by 1,025 kilograms per meter cubed multiplied by 2.4 times 10 to the minus four meters squared multiplied by one second.

Before we go any further, let’s quickly take a look at the units involved in this calculation. In the numerator, we have kilograms. And in the denominator, we have kilograms per meter cubed multiplied by meters squared multiplied by seconds. We can simplify this by multiplying both the numerator and the denominator by meters cubed, where the meters cubed in the denominator will cancel with the meters cubed in the kilograms per meter cubed. Then, we can see that the kilograms in the numerator and the denominator will also cancel.

Next, if we remember that meters cubed is equal to meters times meters times meters and meters squared is equal to meters timesed by meters, then we can cancel two of the meters terms from the numerator and the denominator, finally giving us units of meters per second, which is exactly what we would expect for a speed. Evaluating our expression for the speed of the water, we get 𝑉 is equal to 6.51 meters per second.

Alright, now we’ve calculated the speed of the water flowing through the hose, we can use it with our known value of the length of the hose to work out the amount of time it takes for water to completely pass through the hose. Starting with our equation of 𝑇 sub ℎ is equal to 𝐿 divided by 𝑉, we can substitute our known values of 𝐿 and 𝑉 into this, giving us 𝑇 sub ℎ is equal to 15 meters divided by 6.51 meters per second. Evaluating this, we get 𝑇 sub ℎ is equal to 2.3 seconds to one decimal place. And this is how long water takes to completely pass through the hose.

So the time interval between water entering one end of the hose and leaving the opposite end is 2.3 seconds.