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Chapter 1: Problem 45

During heavy rain, a rectangular section of a mountainside measuring \(2.5 \mathrm{~km}\) wide (horizontally), \(0.80 \mathrm{~km}\) long (up along the slope), and \(2.0 \mathrm{~m}\) deep suddenly slips into a valley in a mud slide. Assume that the mud ends up uniformly distributed over a valley section measuring \(0.40 \mathrm{~km} \times 0.40 \mathrm{~km}\) and that the mass of a cubic meter of mud is \(1900 \mathrm{~kg}\). What is the mass of the mud sitting above an area of \(4.0 \mathrm{~m}^{2}\) in that section?

Short Answer

Expert verified
The mass of the mud above a 4.0 m虏 area is 190,000 kg.

Step by step solution

01

Calculate the volume of the mudslide

First, determine the volume of the rectangular section of the mountainside that slipped. Use the formula for the volume of a rectangular prism: \[ V = \text{width} \times \text{length} \times \text{depth} \]Given:\[ \text{width} = 2.5 \text{ km} = 2500 \text{ m} \]\[ \text{length} = 0.80 \text{ km} = 800 \text{ m} \]\[ \text{depth} = 2.0 \text{ m} \]Thus:\[ V = 2500 \text{ m} \times 800 \text{ m} \times 2.0 \text{ m} = 4,000,000 \text{ m}^3 \]
02

Calculate the volume of mud spread over the valley section

Next, calculate the total volume of mud redistributed in the valley. Use the area of the valley section and the fact that the mud is spread uniformly to deduce how deep the layer of mud becomes. Since we are not given the depth but know the total volume and area, use:\[ \text{Volume of mud} = \text{Area of valley section} \times \text{Depth of mud} \]Given:\[ \text{Area of valley section} = 0.40 \text{ km} \times 0.40 \text{ km} = 0.16 \text{ km}^2 = 160,000 \text{ m}^2 \]
03

Calculate the depth of the mud layer in the valley

Rearrange the volume formula to solve for the depth of mud:\[ \text{Depth of mud} = \frac{\text{Volume of mud}}{\text{Area of valley section}} \]Using the values from step 1 and step 2:\[ \text{Depth of mud} = \frac{4,000,000 \text{ m}^3}{160,000 \text{ m}^2} = 25 \text{ m} \]
04

Determine the volume of mud above a specified area

Now calculate the volume of mud sitting above a 4.0 m虏 area in the valley. Since the depth of mud is uniform, the volume can be calculated as:\[ \text{Volume above 4.0 m}^2 = 4.0 \text{ m}^2 \times 25 \text{ m} = 100 \text{ m}^3 \]
05

Calculate the mass of the mud above the specified area

Using the density of mud (given as 1900 kg/m鲁), calculate the mass of the mud:\[ \text{Mass} = \text{Volume} \times \text{Density} \]\[ \text{Mass} = 100 \text{ m}^3 \times 1900 \text{ kg/m}^3 = 190,000 \text{ kg} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Volume Calculation
Volume calculation is a fundamental concept in physics, especially when dealing with three-dimensional objects. To find the volume of the mudslide in this exercise, we use the formula for the volume of a rectangular prism: \( V = \text{width} \times \text{length} \times \text{depth} \). By converting the given dimensions from kilometers to meters, we get:
  • Width: 2.5 km = 2500 m
  • Length: 0.80 km = 800 m
  • Depth: 2.0 m
The calculated volume is: \( V = 2500 \text{ m} \times 800 \text{ m} \times 2.0 \text{ m} = 4,000,000 \text{ m}^3 \). This large number represents the space the mud occupies, highlighting the power of volume calculation in understanding physical scenarios.
Density
Density is defined as mass per unit volume and is a crucial property in physics. In this problem, the density of the mud is given as 1900 kg/m鲁. This means each cubic meter of mud weighs 1900 kg. Understanding density helps us relate the volume of an object to its mass. This relationship is especially important in the last step of the solution where we convert the volume of mud to its mass using the formula: \( \text{Mass} = \text{Volume} \times \text{Density} \). Always remember that higher density means more mass in the same volume.
Mass Calculation
Mass is a measure of the amount of matter in an object. In this exercise, to find the mass of the mud spread over a specific area, we first determine the volume of the mud, which is influenced by its uniform depth. The key formula here is: \( \text{Mass} = \text{Volume} \times \text{Density} \). After finding the volume of mud over a 4.0 m虏 area to be 100 m鲁, we calculate the mass as: \( 100 \text{ m}^3 \times 1900 \text{ kg/m}^3 = 190,000 \text{ kg} \). Understanding this calculation helps demonstrate how mass depends directly on both volume and density.
Rectangular Prism
A rectangular prism is a three-dimensional geometrical shape with six faces, all of which are rectangles. The problem initially presents a rectangular section of a mountainside that slips into a valley. To find the volume of this prism, we use the formula: \( V = \text{width} \times \text{length} \times \text{depth} \). By plugging in the values, we determine the prism's volume to be 4,000,000 m鲁. Recognizing such shapes and understanding their volume calculations are essential skills in solving similar physics and geometry problems.
Unit Conversion
Unit conversion is often necessary when working with measurements in physics. In this exercise, we convert lengths from kilometers to meters to maintain consistency in our calculations. For example:
  • 2.5 km becomes 2500 m
  • 0.80 km becomes 800 m
  • 0.40 km becomes 400 m
These conversions ensure that all units match up, which is crucial for accurate volume and subsequent calculations. Always double-check unit consistency to avoid errors in your computations.

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Most popular questions from this chapter

Dose We know from our dimensional analysis that if an object maintains its shape but changes its size, its area changes as the square of its length and its volume changes as the cube of its length. Suppose you are a parent and your child is sick and has to take some medicine. You have taken this medicine previously and you know its dose for you. You are \(5^{\prime} 10^{\prime \prime}\) tall and weigh \(180 \mathrm{lb}\), and your child is \(2^{\prime} 11^{\prime \prime}\) tall and wcighs \(30 \mathrm{lb}\). Estimate an appropriate dosage for your child's medicine in the following cases. Be sure to discuss your reasoning. (a) The medicine is one that will enter the child's bloodstream and reach cvery cell in the body. Your dose is \(250 \mathrm{mg}\). (b) The medicine is one that is meant to coat the child's throat. Your dose is \(15 \mathrm{ml}\).

A person on a diet might lose \(2.3 \mathrm{~kg}\) per week. Express the mass loss rate in milligrams per second, as if the dieter could sense the second- by-second loss.

Fermi Physicist Enrico Fermi once pointed out that a standard lecture period (50 min) is close to 1 microcentury. (a) How long is a microcentury in minutes? (b) Using percentage difference \(=\left(\frac{\text { actual }-\text { approximation }}{\text { actual }}\right) 100\) find the percentage difference from Fermi's approximation.

Here are two related problems - one precise, one an estimation. (a) A sculptor builds model for a statue 0 . a terrapin to replace Testudo." She discov. ers that to cast her small scale model she needs \(2 \mathrm{~kg}\) of bronze When she is done, she finds that she can give it two coats of finish. ing polyurethane varnish using exactly one small can of varnish. The final statue is supposed to be 5 times as large as the model in cach dimension. How much bronze will she need? How much varnish should she buy? (Hint: If this seems difficult, you might start by writing a simpler question that is easier to work on before tackling this one.) (b) The human brain has 1000 times the surface area of a mouse's brain. The human brain is convoluted, the mouse's is not. How much of this factor is due just to size (the human brain is bigger)? How sensitive is your result to your estimations of the approximate dimensions of a human brain and a mouse brain?

The micrometer \((1 \mu \mathrm{m})\) is often called the micron. (a) How many microns make up \(1.0 \mathrm{~km}\) ? (b) What fraction of a centimeter equals \(1.0 \mu \mathrm{m}\) ? (c) How many microns are in \(1.0 \mathrm{yd}\) ?
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