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Solutions for Session 7 Homework
See solutions for Problems: H1 | H2 | H3 | H4 | H5
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Problem H1 | |
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Here is the completed table. One formula is L = 2,000(0.6)m.
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Mirror number | | Reflected light (lumens) |
0 | | 2,000 |
1 | | 1,200 |
2 | | 720 |
3 | | 432 |
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<< back to Problem H1
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Problem H2 | |
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The best way to do this, without using more advanced math such as logarithms, is to continue following the table. This is particularly easy with a spreadsheet. The 10th mirror will reflect about 12 lumens.
<< back to Problem H2
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Problem H3 | |
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We'd be looking for the mirror that reflects closest to 200 lumens, which is the 5th mirror (156 lumens). The 4th mirror reflects 259 lumens; this does not depend on the starting amount of light. This means that the 6th mirror reflects about one-tenth the light of the first mirror, the 7th mirror reflects one-tenth the light of the 2nd mirror, and so on.
<< back to Problem H3
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Problem H4 | |
a. | The initial intensity is 1,000 lumens. |
b. | The 1st mirror reflects 700 lumens. |
c. | This fraction is 700 / 1000 = 7/10, or 70 percent. |
d. | Yes, the 2nd mirror appears to reflect about 7/10 of 700 lumens, which is 490 lumens. The 3rd mirror reflects 7/10 of 490 lumens, which is 343 lumens. This pattern appears to continue indefinitely at a constant ratio, so it is an exponential decay situation. |
<< back to Problem H4
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Problem H5 | |
a. | The outputs keep getting smaller, but remain positive, because at each stage a positive number is divided by 5. |
b. | This is an exponential decay function, because successive outputs are getting smaller, and the base is between 0 and 1. |
c. | Zero will never be an output, even though the outputs will become increasingly close to zero. This happens because the numerator remains 1, no matter what the value of x is, while the denominator becomes increasingly larger as x increases. |
d. | Because two positive numbers are used in the division, a negative number can never result. |
e. | In Problems H1-H4, this implies that even after 100 or more mirrors, some light will still be reflected, although the amount of light reflected will become increasingly small. |
<< back to Problem H5
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