Simple Unit Conversion Word Problem Help

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The discussion revolves around a homework problem involving a mudslide and unit conversion to find the mass of mud over a specific area. The initial approach involved calculating the volume of mud from the mountainside and determining its depth in the valley. A mistake was made regarding the dimensions of the area, which was clarified to be 4.0 m², not a square with side lengths of 4.0 m. After correcting the misunderstanding, the final mass of the mud was confirmed to be 1.9 x 10^5 kg, aligning with the textbook answer. The conversation emphasizes the importance of understanding area measurements in solving such problems.
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Homework Statement


During a heavy rain, a section of mountainside measuring 2.5 km horizontally, 0.80 km up along the slope, and 2.0 m deep slips into a valley in a mud slide. Assume that the mud ends up uniformly distributed over a surface area of the valley measuring 0.40km x 0.40km and that mud has a density of ##\frac{1900~kg}{m^3}##. What is the mass of the mud sitting above a 4.0 ##m^2## area of the valley floor.

Homework Equations

The Attempt at a Solution


Please see attached image so you can see how I am visualizing the problem.
My thought was to take the volume of the mud initially on the mountainside, and use that to determine how deep the mud is when it has moved to the valley. From this, I can get a specific volume of mud and use dimensional analysis to get a mass of the mud over the given area. My answer is way off the mark.
 

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opus said:
What is the mass of the mud sitting above a 4.0 ##m^2## area of the valley floor.
Note that the area 4.0 m2 does not correspond to a square of side length 4.0 m. The area of such a square would be 16 m2.
 
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TSny said:
Note that the area 4.0 m2 does not correspond to a square of side length 4.0 m. The area of such a square would be 16 m2.
Ah! That was a dumb mistake. So then the side lengths are each 2m, and my final answer is ##1.9⋅10^5## kg which matches with the back of the book. Thank you!
 
Looks good. Of course, you would still get the same answer even if the base area did not have the shape of a square. All that matters is that it have an area of 4.0 m2.
 
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I'll keep that in mind. Been doing a lot of these to prepare for the start of the semester on Monday, so I appreciate the response at this time of night.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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