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olasank [31]
2 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]2 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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A cup of hot coffee can be cooled by placing a cold spoon in it. A spoon of which of the following materials would be most effec
arsen [322]

Answer: copper

Explanation: when a spoon is placed in a hot coffee, the coffee will cool off based on the concept of heat transfer.

The coffee is at a higher temperature compared to the spoon, hence there is conduction of heat by the spoon from the coffee.

Heat is now transfered from the coffee to the spoon, the rate of heat transfer ( for this case coffee to spoon) is dependent on some properties but for this case, the thermal conductivity is of the most important.

A material of high thermal conductivity have a very fast rate of heat transfer and a material of low thermal conductivity have a slow rate of heat transfer.

Below is the value of thermal conductivity for copper, aluminum, iron and glass.

Copper (385 w/m.K)

Aluminum (205 W/m.K)

Iron (79.5 W/m.K)

Glass (0.8 W/m.K)

It can been seen that copper has the highest value for thermal conductivity which implies that it will conduct heat faster compared to others thus cooling the coffee faster.

8 0
2 years ago
A satellite with a mass of 5.6 E 5 kg is orbiting the Earth in a circular path. Determine the satellite's velocity if it is orbi
solniwko [45]

Answer:

7500 m/s

Explanation:

Centripetal acceleration = gravity

v² / r = GM / r²

v = √(GM / r)

Given:

G = 6.67×10⁻¹¹ m³/kg/s²

M = 5.98×10²⁴ kg

r = 6.8×10⁵ + 6.357×10⁶ = 7.037×10⁶ m

v = √(6.67×10⁻¹¹ (5.98×10²⁴) / (7.037×10⁶))

v = 7500

The orbital velocity is 7500 m/s.

7 0
1 year ago
Platinum (pt) has the fcc crystal structure, an atomic radius of 0.1387 nm, and an atomic weight of 195.08 g/mol. what is its th
prohojiy [21]
The equation to be used is written as:

ρ = nA/VcNₐ
where
ρ is the density
n is the number of atoms in unit cell (for FCC, n=4)
A is the atomic weight
Vc is the volume of the cubic cell which is equal to a³, such that a is the side length (for FCC, a = 4r/√2, where r is the radis)\
Nₐ is Avogradro's constant equal to 6.022×10²³ atoms/mol

r = 0.1387 nm*(10⁻⁹ m/nm)*(100 cm/1m) = 1.387×10⁻⁸ cm
a = 4(1.387×10⁻⁸ cm)/√2 = 3.923×10⁻⁸ cm
V = a³ = (3.923×10⁻⁸ cm)³ = 6.0376×10⁻²³ cm³

ρ = [(4 atoms)(195.08 g/mol)]/[(6.0376×10⁻²³ cm³)(6.022×10²³ atoms/mol)]
ρ = 21.46 g/cm³
7 0
1 year ago
Read 2 more answers
Two runners ran side by side each holding one end of a horizontal pole. What would most likely happen if one of the runners bega
Anettt [7]

Answer:

See explanation.

Explanation:

If each runner was holding the pole, the runner in the water side of the pole would probably be behind the other runner. Since running in knee deep is hard and makes you slower, the pole would be slanted.

5 0
1 year ago
In a college homecoming competition, eighteen students lift a sports car. While holding the car off the ground, each student exe
Nata [24]

Answer:

Explanation:

Given

Each student exert a force of F=400 N

Let mass of car be m

there are 18 students who lifts the car

Total force by 18 students F=18\times 400=7200 N

therefore weight of car W=7200

mass of car m=\frac{W}{g}

m=\frac{7200}{9.8}=734.69 kg

(b)7200 N \approx 1618.624\ Pound-force

734.69 kg\approx 1619.71 Pounds                  

6 0
1 year ago
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