Simple Unit Conversion Word Problem Help

In summary, the problem involves a mud slide where 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. The mud ends up uniformly distributed over a surface area of the valley measuring 0.40 km x 0.40 km and has a density of ##\frac{1900~kg}{m^3}##. The task is to find the mass of the mud above a 4.0 ##m^2## area of the valley floor. Using dimensional analysis and considering the shape of the given area, the mass is calculated to be ##1.9⋅10^5## kg.
  • #1
opus
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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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  • #2
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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  • #3
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!
 
  • #4
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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  • #5
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.
 

What is a simple unit conversion word problem?

A simple unit conversion word problem involves converting a quantity from one unit of measurement to another. For example, converting 10 feet to meters or 2 pounds to kilograms.

Why are unit conversions important?

Unit conversions are important because different countries and industries use different units of measurement. Being able to convert between units allows for better communication and understanding between individuals and organizations.

How do I solve a simple unit conversion word problem?

To solve a simple unit conversion word problem, you need to know the conversion factor between the two units and use it to convert the given quantity. You can also use unit conversion tables or online conversion calculators to help with the process.

What are some common units of measurement used in unit conversion word problems?

Common units of measurement used in unit conversion word problems include length (meters, feet, inches), mass (kilograms, pounds, ounces), volume (liters, gallons, cubic feet), and time (seconds, minutes, hours).

Can you provide an example of a simple unit conversion word problem?

Sure, here's an example: Convert 5 miles to kilometers. The conversion factor between miles and kilometers is 1 mile = 1.60934 kilometers. Therefore, 5 miles = 5 x 1.60934 = 8.0467 kilometers.

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