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tester [92]
4 years ago
12

Which of the following is a use of radioactive isotopes? Question 7 options: a. can determine the ages of rocks and fossils b. c

an be used to treat cancer and kill bacteria that cause food to spoil c. can be used as "tracers" to follow the movements of substances within organisms d. all of the above
Physics
1 answer:
Zielflug [23.3K]4 years ago
8 0

Answer:

d. all of the above

Explanation:

a) Absolute dating technique is used to determine the age of rocks and fossil fuels, by the use of radioactive elements, such as, Carbon-14. The process involves the measurement of life of fossils through the remaining amount of radioactive element. It is possible by the knowledge of half life of radioactive element.

b) The radioactive isotope of Cobalt (Cobalt-60) is used to treat localized cancer. Other radioactive elements are also used to kill bacteria and treat cancer, such as, Iodine-131 is used to treat thyroid cancer

c) Radioactive elements are also used as tracers, such as Sodium-24, which is used as a tracer in blood. Other radioisotopes of Hydrogen, Carbon. Phosphorus and Sulfur, etc, are also used as tracers in different chemical reactions.

<u>Thus, all the given options are correct and the correct option is all of the above (d).</u>

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Two point charges are fixed on the y axis a negative point charge q1-26 μC at y1 = +0.21 m and a positive point charge q2 art y2
11111nata11111 [884]

Answer:

3.13\times 10^{-5} C

Explanation:

We are given that

q_1=26\mu C=26\times 10^{-6} C

1\mu C=10^{-6} C

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q=8.1\mu C=8.1\times 10^{-6} C

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We have to find the magnitude of q2.

We know that

F=\frac{kq_1q_2}{r^2}

Where k=9\times 10^9

Using the formula

Force on charge q due to charge q1

F_1=\frac{9\times 10^9\times 26\times 10^{-6}\times 8.1\times 10^{-6}}{(0.21)^2}

F_1=42.98 N

Force on charge q due to point charge q2

F_2=\frac{9\times 10^9\times q_2\times 8.1\times 10^{-6}}{(0.39)^2}

F_2=4.79\times 10^5 q_2

F=F_1-F_2

28=42.98-4.79\times 10^5q_2

4.79\times 10^5q_2=42.98-28=14.98

q_2=\frac{14.98}{4.79\times 10^5}=3.13\times 10^{-5} C

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