Move the horizontal and vertical sliders to stretch the image of the triangle. When you've settled on a pair of transformed imag
es, record the coordinates of the images in the table. Be sure to list coordinates in both the horizontal (x) and vertical (y) directions. Also record the scale factors, nx and ny. Do this at least three times to fill the table.
ANSWER:This answer assumes that nx and ny belong to the set {0.5, 1, 1.5}.
nx A A′ B B′ C C′
0.5 (2, 1) (1, 1) (4, 2) (2, 2) (5, 1) (2.5, 1)
1 (2, 1) (2, 1) (4, 2) (4, 2) (5, 1) (5, 1)
1.5 (2, 1) (3, 1) (4, 2) (6, 2) (5, 1) (7.5, 1)
ny A A′ B B′ C C′
0.5 (2, 1) (2, 0.5) (4, 2) (4, 1) (5, 1) (5, 0.5)
1 (2, 1) (2, 1) (4, 2) (4, 2) (5, 1) (5, 1)
1.5 (2, 1) (2, 1.5) (4, 2) (4, 3) (5, 1) (5, 1.5)
ANSWER:This answer assumes that nx and ny belong to the set {0.5, 1, 1.5}.
nx A A′ B B′ C C′
0.5 (2, 1) (1, 1) (4, 2) (2, 2) (5, 1) (2.5, 1)
1 (2, 1) (2, 1) (4, 2) (4, 2) (5, 1) (5, 1)
1.5 (2, 1) (3, 1) (4, 2) (6, 2) (5, 1) (7.5, 1)
ny A A′ B B′ C C′
0.5 (2, 1) (2, 0.5) (4, 2) (4, 1) (5, 1) (5, 0.5)
1 (2, 1) (2, 1) (4, 2) (4, 2) (5, 1) (5, 1)
1.5 (2, 1) (2, 1.5) (4, 2) (4, 3) (5, 1) (5, 1.5)
1 answer:
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Answer:
The approximate difference in the half-lives of the isotopes is 66 days.
Step-by-step explanation:
The decay of an isotope is represented by the following differential equation:
Where:
- Current mass of the isotope, measured in kilograms.
- Time, measured in days.
- Time constant, measured in days.
The solution of the differential equation is:
Where is the initial mass of the isotope, measure in kilograms.
Now, the time constant is cleared:
The half-life of a isotope () as a function of time constant is:
The half-life difference between isotope B and isotope A is:
If ,
and
, the difference in the half-lives of the isotopes is:
The approximate difference in the half-lives of the isotopes is 66 days.