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olasank [31]
3 years ago
5

Which of the following scenarios would be optimal for obtaining a date from radioactive decay using these isotopes: 87Rb, 147Sm,

235U, 238U, 40K, or 14C? There may be more than one answer that is appropriate. Explain your reasoning for why the remaining scenario(s) would be inappropriate/impossible to use that particular isotope. Answers should include a discussion on usable ages for each system and whether the necessary isotopes would be found in the material to be dated.
a. A meteorite that formed early in the formation of the solar system.

b. A rock formed through a mountain building event around 420 million years ago.

c. Volcanic ash from an eruption 60 million years ago.

d. An earthquake scarp that formed along the San Andreas Fault 50 years ago.

e. An Incan archaeological dig site in the highlands of Peru.

f. A tree from a forest in England that is suspected to be the oldest in the British Isles.
Physics
1 answer:
REY [17]3 years ago
8 0

Answer:

a) 238U, 40K and 87Rb, b)   235U and to a lesser extent 40K , c)  he 235U,

d) possibility is 14C , e)this period would be ideal for 14C , f) 14C should be used since it is the one with the least average life time, even though the measurements must be very careful

Explanation:

One of the applications of radioactive decay is the dating of different systems.

To do this, the quantity of radioactive material in a meter is determined and with the average life time, the time of the sample is found.

Let's write the half-life times of the given materials

87Rb T ½ = 4.75 1010 years

147Sm T ½ = 1.06 1011 years

235U = 7,038 108 years

238U = 4.47 109 years

40K = 1,248 109 years

14C = 5,568 103 years

we already have the half-life of the different elements given

a) meteors. As these decomposed in the formation of the solar system, their life time is around 3 109 to 5 109 years, so it is necessary to look for elements that have a life time of this order, among the candidates we have 238U, 40K and 87Rb if these elements were at the moment of the formation of these meteors, there must still be rations in them, instead elements 14C already completely adequate

b) rock. The formation period is 4.20-108 years, therefore one of the most promising elements is 235U and to a lesser extent 40K since it is more abundant in rocks. The other elements with higher life times have not decayed and therefore will not give a true value and the 14C is completely decayed

c) volcanic ash. Formation time 6107 years, the only element that has the possibility of having a count is the 235U, the others have a life time so long that they have not decayed and the 14C is complete, unbent

d) scarp of an earthquake formation time 5 101 years, The only one that has any possibility is 14C even when it has declined very little, all the others, you have time to long that has not decayed

e) INCA excavation. The time of this civilization is about 10000 to 500 years (104 to 5 102 years), we see that this period would be ideal for 14C since it has some period of cementation, the others have not decayed

f) Tree in Blepharitis. 14C should be used since it is the one with the least average life time, even though the measurements must be very careful because of a period of disintegration. We have such a long time that they have not decayed

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morpeh [17]
A is the correct answer
7 0
3 years ago
3 small identical balls have charges -3×10^-12 ,8×10^-12 and 4×10^-12 find the charge​
Norma-Jean [14]

Answer:

  3×10^-12 C  

Explanation:

The total of the three charges is ...

   (-3 +8 +4)×10^-12 C = 9×10^-12 C

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2 years ago
A displacement vector 5 units long makes an angle of 67° with the x-axis. What is the horizontal component of displacement?
Harman [31]

Yo sup??

magnitude of original vector is 5 units.

angle made with x axis is 67°

horizontal component=5*cos67°

=5*0.4

=2 units (approx)

Therefore the correct answer is option B

Hope this helps

7 0
3 years ago
Read 2 more answers
a projectile is fired in such away that its horizontal range is equal to three times its naximum height.what is the angle of pro
finlep [7]

Answer:

\theta=53.13^o

Explanation:

<u>2-D Projectile Motion</u>

In 2-D motion, there are two separate components of the acceleration, velocity and displacement. The horizontal component has zero acceleration, while the acceleration in the vertical direction is always the acceleration due to gravity. The basic formulas for this type of movement are

V_x=V_{ox}=V_ocos\theta

V_y=V_{oy}-gt=V_osin\theta-gt

x=V_{ox}t

\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}

\displaystyle x_{max}=\frac{2V_{ox}V_{oy}}{g}

\displaystyle y_{max}=\frac{V_{oy}^2}{2g}

The projectile is fired in such a way that its horizontal range is equal to three times its maximum height. We need to find the angle \theta at which the object should be launched. The range is the maximum horizontal distance reached by the projectile, so we establish the base condition:

x_{max}=3y_{max}

\displaystyle \frac{2V_{ox}V_{oy}}{g}=3\frac{V_{oy}^2}{2g}

Using the formulas for V_{ox}, V_{oy}:

\displaystyle \frac{2V_{o}cos\theta V_{o}sin\theta}{g}=3\frac{V_{o}^2sin^2\theta}{2g}

Simplifying

4cos\theta sin\theta=3sin^2\theta

Dividing by sin\theta

4cos\theta=3sin\theta

Rearranging

tan\theta=\frac{4}{3}

\theta=arctan\frac{4}{3}

\theta=53.13^o

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