A 6.0 m ✕ 12.0 m swimming pool slopes linearly from a 1.0 m depth at one end to a 3.0 m depth at the other. What is the mass of water in the pool?

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Accepted Answer

Hi everyone. In this practice problem, we're being asked to determine the mass of water contained in a structure. We will have an 8 m wide and 60 m long water retention structure with varying depths. The shallow end is 2 m deep. While the deep end reaches 5 m which rises linearly, we were being asked to determine the mass of water contained in the structure. And the options given are a 3.49 times 10 to the power of kg. B 5.21 times 10 to the power of 5 kg. C 3.68 times 10 to the power of 5 kg and D 4.48 times 10 to the power of 5 kg. So in this particular practice problem, let's start with actually um drawing the top and side view of the container. So the top view is just essentially going to be a rectangle like. So which in this case, the uh white, the width is going to be 8 m and the length is going to be m just like. So the difference within the top few, I'm just gonna write that as top and the side view is that for the side view, we have essentially uh trap trapezoid because the uh depth of the container will rise linearly. So in this case, this is going to be the side view and the 16 m uh length is going to stay constant or the same. And the depth is going to have to be 2 m on this side. And on the other side, it's going to be 5 m and it will rise linearly. So in this case, the mass of water, which is what we ask in this problem, M water can be calculated by multiplying the density of water row water multiplied by the volume, the total volume of the container. So M will equals to row multiplied by V. So the first thing that we wanna do is to calculate the volume. So volume will be equals to area multiplied by width. So in this case, the area is going to be the area of the trapezoid. So a equals a of trapezoid. And if you recall the formula to calculate the area for the trapezoid, we can calculate it by A will be equals to half multiplied by the first uh the first site which is 2 m plus the second side, which is 5 m. So the uh summation of the two sides that are parallel to one another divided by two multiplied by the length which in this case is going to be m. So calculating the area, the area will then come up to a value of 56 m squared. Thus, we can calculate the volume where the volume will then be equal to 56 m squared multiplied by the width which is going to be 8 m which will come out to a volume of 448 m squared. Finally, we can actually uh substitute our volume into the mass equation where mass will equals to row multiplied by volume. The row is 1000 kg per meter cube for water multiplied that with the volume which is 448 m squared. And then calculating this, we will get the mass to be 448, kg or 4.48 times 10 to the power of 5 kg. So there we have the final answer to this practice problem. The total mass of the water in the container is 4.48 times 10 to the power of 5 kg, which will correspond to option D in our answer choices. So option D will be the answer to this particular practice problem. And that will be it for this video. If you guys still have any sort of confusion, please feel free to check out our adolescent videos on similar topics available on our website. But other than that, that will be it for this video. Thank you.

A 6.0 m ✕ 12.0 m swimming pool slopes linearly from a 1.0 m depth at one end to a 3.0 m depth at the other. What is the mass of water in the pool?

Verified Solution

Key Concepts

Volume of a Trapezoidal Prism

Density of Water

Mass Calculation