Transparency of aLake Environmental sci-entists measure the inten-sity of light at variousdepths in a lake to find the"transparency" of thewater. Certain levels oftransparency are requiredfor the biodiversity of thesubmerged macrophytepopulation. In a certainlake the intensity of lightat depth x is given byI = 10e-0.008xwhere I is measured in lumens and x in feet.(a) Find the intensity I at a depth of 30 ft.(b) At what depth has the light intensity dropped to I = 5?

Substitution of the respective values into the given function helps to find the corresponding…

Answered: Transparency of a Lake Environmental…

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter4: Polynomial And Rational Functions

Section4.6: Variation

Problem 31E

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The Ideal Gas Law A mole of a chemical is a fixed number, like a dozen, of molecules or atoms in the case of an element of that compound. A mole of water, for example, is about 18 grams, or just over half an ounce in your kitchen. Chemists often use the mole as the measure of the amount of a chemical compound. A mole of carbon dioxide has a fixed mass, but the volume V that it occupies depends on pressure p and temperature T; greater pressure tends to compress the gas into a smaller volume, whereas increasing temperature tends to make the gas expand into a larger volume. If we measure the pressure in atmospheres 1 atm is the pressure exerted by the atmosphere at sea level, the temperature in kelvins, and the volume in liters, then the relationship is given by the ideal gas law: pV=0.082T. a. Solve the ideal gas law for the volume V. b. What is the volume of 1 mole of carbon dioxide under 3 atm of pressure at a temperature of 300 kelvins? c. Solve the ideal gas law for pressure. d. What is the pressure on 1 mole of carbon dioxide if it occupies a volume of 0.4 liter at a temperature of 350 kelvins? e. Solve the ideal gas law for temperature. f. At what temperature will 1 mole of carbon dioxide occupy a volume of 2 liters under a pressure of 0.3 atm?
Decay of Litter Litter such as leaves falls to the forest floor, where the action of insects and bacteria initiates the decay process. Let A be the amount of litter present, in grams per square meter, as a function of time t in years. If the litter falls at a constant rate of L grams per square meter per year, and if it decays at a constant proportional rate of k per year, then the limiting value of A is R=L/k. For this exercise and the next, we suppose that at time t=0, the forest floor is clear of litter. a. If D is the difference between the limiting value and A, so that D=RA, then D is an exponential function of time. Find the initial value of D in terms of R. b. The yearly decay factor for D is ek. Find a formula for D in term of R and k. Reminder:(ab)c=abc. c. Explain why A=RRekt.
Maximum Sales Growth This is a continuation of Exercise 10. In this exercise, we determine how the sales level that gives the maximum growth rate is related to the limit on sales. Assume, as above, that the constant of proportionality is 0.3, but now suppose that sales grow to a level of 4 thousand dollars in the limit. a. Write an equation that shows the proportionality relation for G. b. On the basis of the equation from part a, make a graph of G as a function of s. c. At what sales level is the growth rate as large as possible? d. Replace the limit of 4 thousand dollars with another number, and find at what sales level the growth rate is as large as possible. What is the relationship between the limit and the sales level that gives the largest growth rate? Does this relationship change if the proportionality constant is changed? e. Use your answers in part d to explain how to determine the limit if we are given sales data showing the sales up to a point where the growth rate begins to decrease.
Radius of a Shock Wave An explosion produces a spherical shock wave whose radius R expands rapidly. The rate of expansion depends on the energy E of the explosion and the elapsed time t since the explosion. For many explosions, the relation is approximated closely by R=4.16E0.2t0.4. Here R is the radius in centimeters, E is the energy in ergs, and t is the elapsed time in seconds. The relation is valid only for very brief periods of time, perhaps a second or so in duration. a. An explosion of 50 pounds of TNT produces an energy of about 1015 ergs. See Figure 2.71. How long is required for the shock wave to reach a point 40 meters 4000 centimeters away? b. A nuclear explosion releases much more energy than conventional explosions. A small nuclear device of yield 1 kiloton releases approximately 91020 ergs. How long would it take for the shock wave from such an explosion to reach a point 40 meters away? c. The shock wave from a certain explosion reaches a point 50 meters away in 1.2 seconds. How much energy was released by the explosion? The values of E in parts a and b may help you set an appropriate window. Note: In 1947, the government released film of the first nuclear explosion in 1945, but the yield of the explosion remained classified. Sir Geoffrey Taylor used the film to determine the rate of expansion of the shock wave and so was able to publish a scientific paper concluding correctly that the yield was in the 20-kiloton range.

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