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An old campfire is uncovered during an archaeological dig. Its charcoal is found to contain less than 1/1000 the normal amount o

f ^{14}\text{C} 14 C. Estimate the minimum age of the charcoal, noting that

1 answer:

PIT_PIT [12.4K]2 months ago

5 0

During an archaeological excavation, an ancient campfire is uncovered. The charcoal is determined to have significantly less than 1/1000 of the standard amount of $^{14}\text{C}$ . Calculate the minimal age of the charcoal, taking into account that $2^{10} = 1024$

Response:

57300 years

Step-by-step breakdown:

Using the relationship of half-life time against fraction, which can be expressed as:

$\dfrac{N}{N_o} = (\dfrac{1}{2})^{\frac{t}{t_{1/2}}$

In this context,

N indicates the current atom

represents the initial atom $N_o$

t signifies the time

denotes the half-life $t_{1/2}$

Since the charcoal was found to contain less than 1/1000 of the typical amount of $^{14}\text{C}$ .

Thus;

$\dfrac{N}{N_o} = \dfrac{1}{1000}$

However; the objective is to estimate the minimum age of the charcoal while noting $2^{10} = 1024$

this means $2^{10} = 1024$ , then:

$\dfrac{1}{1000}> \dfrac{1}{1024}$

$\dfrac{1}{1000}> \dfrac{1}{2^{10}}$

$\dfrac{1}{1000}> (\dfrac{1}{2})^{10}$

If

$\dfrac{N}{N_o} = \dfrac{1}{1000}$

Then

$\dfrac{N}{N_o} > (\dfrac{1}{2})^{10}$

Consequently, it can be estimated that the minimum time elapsed is 10 half-lives.

For $^{14}\text{C}$ , the standard half-life time is 5730 years

Thus, the estimation of the minimum age of the charcoal is 5730 years × 10

= 57300 years

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I WILL AWARD BRAINLIEST!! PLEASE HELP!!! The figure below shows the movement of a pedestrian from point B to point E. Using the

Inessa [12570]

A) The speed of pedestrian BC is 5 km/h

The speed of pedestrian CD is 0 km/h

The speed of pedestrian DE is 5 km/h

B) He reached E since the stop after 6 hours

C) The formula for section BC is d(t) = 40 - 5t

The formula for section CD is d(t) = 20

The formula for section DE is d(t) = 50 - 5t

Step-by-step breakdown:

A)

In the time-distance graph, speed represents the change in distance with respect to time

this means speed = Δd/Δt ⇒ (slope of the line)

For line BC:

1. Δd = 40 - 20 = 20 km

2. Δt = 4 - 0 = 4 hours

3. The speed = 20 ÷ 4 = 5 km/h

The speed of pedestrian BC stands at 5 km/h

For line CD:

1. Δd = 20 - 20 = 0 km

2. Δt = 6 - 4 = 2 hours

3. The speed = 0 ÷ 2 = 0 km/h

The speed of pedestrian CD is 0 km/h

For line DE:

1. Δd = 20 - 0 = 20 km

2. Δt = 10 - 6 = 4 hours

3. The speed = 20 ÷ 4 = 5 km/h

The speed for pedestrian DE is 5 km/h

B)

∵ He stopped at t = 4 hours

∵ He reached point E at t = 10 hours

∵ 10 - 4 = 6 hours

He arrived at E since the stop took 6 hours

C)

<pthe line="" equations="" are="" characterized="" by:=""><pthe general="" form="" of="" a="" line="" equation="" is="" f="" mx="" c="" where="" m="" represents=""><pthe slope="" and="" c="" is="" the="" y-intercept="" value="" of="" y="" when="" x="" equals="">

1. f(x) is referenced as d(t)

2. m is the speed

3. x corresponds to t

4. You can calculate c by substituting any coordinates of a point along the line into the formula

<pline bc="">

Line BC has a negative slope because d decreases as t increases

∵ m = -5 and c = 40

Thus, d(t) = 40 - 5t

The equation for section BC is d(t) = 40 - 5t

Line CD

Line CD is a horizontal line (which follows the form of any horizontal line represented as y = c)

Thus, m = 0 and c = 20

Therefore, d(t) = 20

The equation for section CD is d(t) = 20

Line DE

Line DE features a negative slope as d decreases along with an increase in t

∵ m = -5

Thus, d(t) = -5t + c

The value of c can be determined by substituting point D's coordinates into the equation

∵ D's coordinates are (6, 20)

Thus, 20 = -5(6) + c

So, 20 = -30 + c

Adding 30 to both sides yields

∴ c = 50

Thus, d(t) = 50 - 5t

The equation for section DE is d(t) = 50 - 5t

Further Learning:

<padditional information="" regarding="" distance="" speed="" and="" time="" can="" be="" found="" in="">

</padditional></pline></pthe></pthe></pthe>

3 0

3 months ago

An astronaut on the moon throws a baseball upward. The astronaut is 6 ft, 6 in. tall, and the initial velocity of the ball is 40

lawyer [12517]

Response:

0.14 s

Detailed breakdown:

s = -2.7 t² + 40t + 6.5

Set s = 12

12 = -2.7t² + 40t + 6.5 Rearranging yields

-2.7t² + 40t + 6.5 - 12 = 0

-2.7t² + 40t - 5.5 = 0

Utilize the quadratic formula

$x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$

a = -2.7; b = 40; c = -5.5

$x = \frac{-40\pm\sqrt{40^2 - 4\times (-2.7) \times (-5.5)}} {2(-2.7)}$

$x = \frac{-40\pm\sqrt{1600-59.4}}{-5.4}$

$x = \frac{-40\pm\sqrt{1540.6}}{-5.4}$

$x = \frac{-40\pm 39.25}{-5.4}$

x = 7.41 ± 7.27

x₁ = 0.14; x₂ = 14.68

The graph indicates roots at x₁ = 0.134 and x₂ = 14.68.

The surface of the Moon stands at -12 ft. Thus, the ball will reach a height of 12 ft above the Moon’s surface (crossing the x-axis) at 0.14 s.

The second root indicates when the ball is again 12 ft above the lunar surface as it descends.

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2 months ago

A snake tank measures 1.8 m x 0.5 m x 0.5 m. What is the surface area of the tank including the top? Use the formula: SA = 2hl+2

AnnZ [12381]

Answer:

4.1

Step-by-step explanation:

1.8 multiplied by 0.5 equals 0.9

0.5 multiplied by 0.5 gives 0.25

Calculating 2 times (0.9 plus 0.9 plus 0.25) results in 2 times (1.8 plus 0.25), which is 2 times 2.05

Multiplying 2 by 2.05 results in 4.1

Thus, the result is 4.1

I trust this was useful!

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3 months ago

Suppose you are determining the growth rate of two species of plants. Species A is 12 cm tall and grows 2 cm per month. Species

lawyer [12517]

Species A stands at 12 cm tall, growing at a rate of 2 cm monthly. H(m) = 12 + 2m, while Species B starts at 10 cm tall and increases by 3 cm each month. H(m) = 10 + 3m.

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2 months ago

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