WEBVTT
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45 kilograms of a liquid with a constant density of 1055 kilograms per meter cubed flows smoothly through a 2.5-meter-long pipe each second.
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What is the cross-sectional area of the pipe?
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Letβs start by drawing a diagram.
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We have a pipe that has a length that we will call πΏ and a cross-sectional area that we will call π΄.
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We are told that there is a liquid flowing through the pipe that has a constant density that we will call π.
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Now, if we were to place some sort of container at the end of the pipe to catch the liquid and we were to observe it for a certain amount of time, after a time that we will call capital π, a certain mass of liquid will have collected here in our container.
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And we will call this mass π.
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The question tells us that the pipe is 2.5 meters long.
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So πΏ is equal to 2.5 meters.
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The question also tells us that the liquid has a constant density of 1055 kilograms per meter cubed.
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So π is equal to 1055 kilograms per meter cubed.
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And we are also told that 45 kilograms of the liquid flows through the pipe each second.
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So π is equal to 45 kilograms and π is equal to one second.
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The question asks us to calculate the cross-sectional area of the pipe.
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So we must calculate π΄.
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We will start with the liquidβs mass flow rate, which is the amount of mass of the liquid that has flowed through the pipe per unit of time.
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And this is equal to the density of the liquid multiplied by the cross-sectional area of the pipe it is flowing in multiplied by the liquidβs speed.
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However, we arenβt told the liquidβs speed, so we must calculate it.
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We know that speed is equal to distance traveled divided by the time taken to travel that distance.
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And in this case we are told that the liquid flows through the entire pipe each second.
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So π is equal to πΏ divided by π.
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Substituting this back into our equation, we get π divided by π is equal to ππ΄ multiplied by πΏ divided by π.
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We can simplify this by multiplying both sides of the equation by π.
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And then we see that the πβs on the left and also the right cancel.
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Now, we will rearrange this equation to get an expression for π΄, the cross-sectional area of the pipe.
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We will divide both sides of the equation by π multiplied by πΏ.
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And we see that these πβs on the right cancel and the same for the πΏβs, leaving us with our expression for π΄.
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Writing this a bit more neatly, π΄ is equal to π divided by π multiplied by πΏ.
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Now, we can carry on and substitute our known values of π, π, and πΏ into this equation.
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And we notice that these are all in SI units.
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So we donβt need to convert any of these before continuing.
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So π΄ is equal to 45 kilograms divided by 1055 kilograms per meter cubed multiplied by 2.5 meters.
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Evaluating this expression gives π΄ is equal to 0.017 meters squared to three decimal places.
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The cross-sectional area of the pipe is equal to 0.017 meters squared.