- $99 \textrm{ m}^3\textrm{/s}$
- $9.9 \textrm{ m}^3\textrm{/s}$
- $0.099 \textrm{ m}^3\textrm{/s}$
- $0.0099 \textrm{ m}^3\textrm{/s}$

(d)

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This is College Physics Answers with Shaun Dychko. We have a large container of water that is open to the air which is a clue that means that the pressure at the top of this container is atmospheric pressure. And this container has a hole with an area of ten square centimeters at a 0.5 meters below the surface of the water. And so the question is what is the flow rate of the water coming out of this hole. So what is <i>Q</i>? In other words, the volume flow rate in cubic meters per second. So, we're going to set up Bernoulli's Equation here and we're going to figure out the speed of the water coming out of the hole. And then, we'll multiply that by the hole's cross-sectional area to figure out the volume flow rate. So Bernoulli's Equation at the top of the water here, the surface of the water is pressure at the top which we've already said is atmospheric pressure plus one half times density of water times the speed with which the water is moving downwards at the top here plus <i>rho g</i> times the height of the top and that equals the pressure at the hole which also is atmospheric pressure plus one half rho times the the speed of waters coming out of the hole and we're going to solve for that in a second, plus <i>rho g</i> times the height of the hole. Now, we know that the pressure at the top and the pressure at the hole are both atmospheric pressure since they're exposed to the air and so these terms are the same and so they cancel. And then, we can also say that the speed of the water, this is the speed at which the water is moving downwards since it's leaking out of the hole. So that's speed is approximately zero at the top and we know that because of this clue that says large container. And because the container is really large, that means there's such a huge cross-sectional area here that the water has to only go down a very tiny amount in order to provide a volume of water which is large maybe coming out of this hole, because the hole is so small in comparison to the top of the container. So, we can say that the speed with which the water is going downwards is approximately zero. Now, with that said we can get rid of this term and we're going to solve for kinetic energy of the water from the hole term by saying one half <i>rho Vh</i> squared is <i>rho</i> times <i>g</i>, which is a common factor between this term and this one times the height of the top of the water the surface minus the height of the hole. So I took this term to the other side by subtracting <i>rho gh</i> from both sides here. Then, we multiply both sides by two over density and the density just cancels. It's common factor and then take the square root of both sides, as well. Then you get the speed with which the water is coming out of the hole is a square root of two <i>g</i> times the difference in height between the surface and the hole. And we're told that the difference in height is five meters. And so we'll multiply that by acceleration to the gravity and then times by two and take the square root of that product and we get 9.905 meters per second. And there's a convincing distractor here is this option b is tempting because it's 9.9 and we just got an answer of 9.9 but this is the speed of the water. It is not the volume flow rate of cubic meters per second so that's not our answer. Instead, we have to multiply that speed by the cross-sectional area of the hole to figure out the volume flow rate. And so we're told the hole has an area of ten square centimeters and we'll convert that into square meters by multiplying by one meters for every hundred centimeters twice and this works out to 0.0099 cubic meters per second. And that is answer d.