What is the length of decay of carbon-14?

a.

A couple days

b.

A couple hours

c.

A couple seconds

d.

A few thousand years

Answers

Answer 1

Answer:

My answer will be (D) . A few thousand years

Explanation:

Carbon-14 is a radioactive isotope of carbon, containing 6 protons and 8 neutrons, that is present in the earth's atmosphere in extremely low concentrations.2

It is naturally produced in the atmosphere by cosmic rays (and also artificially by nuclear weapons), and continually decays via nuclear processes into stable nitrogen atoms.

Suppose we have a sample of a substance containing some carbon-14. Let m be the mass of carbon-14 in nanograms after t years. 3

It turns out that, if the sample is isolated, then m and t approximately 4 satisfy the differential equation

dmdt=−0.000121m.

Suppose our sample initially contains 100 nanograms of carbon-14. Let's investigate what happens to the sample over time.

First, we can solve the differential equation. Since m has a continuous decay rate of −0.000121, a general solution to the differential equation is

m(t)=Ce−0.000121t,

where C is a constant. Substituting the initial condition t=0, m=100 gives C=100, so

m(t)=100e−0.000121t.

With this formula, we can calculate the amount m of carbon-14 over the years.

Mass of carbon-14 in sample

t (years) m (ng to 4 decimal places)

0 100.0000

100 98.7973

1000 88.6034

2000 78.5056

5000 54.6074

10 000 29.8197

20 000 8.8922

Every year, the mass m of carbon-14 is multiplied by e−0.000121≈0.999879. After 100 years, 98.7973 nanograms still remain. After 1000 years, we still have 88.6034 nanograms. But after 5000 years, however, almost half of the carbon-14 has decayed.

Graph showing the decay of 100 nanograms of Carbon-14.

Detailed description

half-life of carbon-14  

Example

How long does it take for precisely half of the carbon-14 in the sample to decay; that is, when does m=50? Give the answer to three significant figures.

Solution

The mass of carbon-14 in our sample is given by

m(t)=100e−0.000121t.

So we solve 50=100e−0.000121t, which gives e−0.000121t=12. Hence,

t=loge12−0.000121≈5730 years (to three significant figures).

The time period calculated in this example is called the half-life of carbon-14. In fact over any period of 5730 years, the amount of carbon-14 in an isolated sample will decay by half. This fact is used in radiocarbon dating to determine the age of fossils up to 60 000 years old. Roughly speaking, while an organism is alive, its interactions with its environment maintain a constant ratio of carbon-14 to carbon-12 in the organism; but after it dies, the carbon-14 is no longer replenished, and the ratio of carbon-14 to carbon-12 decays in a predictable way. (See Wikipedia Open new window for more on radiocarbon dating.)

Exercise 1

Explain why the mass of carbon-14 in the sample is given (approximately) by

m(t)=100(12)t5730,

and hence explain why the amount of carbon-14 in the sample decays by half over any period of 5730 years.

Half-life in general

In general, whenever a quantity x(t) obeys an exponential decay equation

x(t)=Cekt,

where the continuous decay rate k is negative, then the quantity x has a half-life T. After any time period of length T, the quantity x decreases by half. Let us see why.

As k is negative, the factor ekt decreases from 1 (at t=0) towards 0 (as t approaches ∞). Therefore there is a time t=T such that

ekT=12.

We now solve for T and obtain

kT=loge12=−loge2,

so

T=−1kloge2.

This T is the half-life. From time t=0 to time t=T, the factor ekt decreases from e0=1 to ekT=12, that is, decreases by half. Similarly, over any time period of length T, the term ekt decreases by half. 5

Note that, when k=−0.000121, we obtain T=5730, in agreement with our calculation for carbon-14.

Hope this work

If I'm wrong I'm sorry

If I'm right thank you (brainliest plz )

Answer 2
the answer is D a few thousand years!!!!

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Answers

The primary medium through which the sound waves travel is

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Answers

Answer:

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Explanation:

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Model = VSO Ha

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Answers

Answer:

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Answers

Answer:

f = 2445 Hz

Explanation:

i did the quiz

The frequency that the driver (observer) detects is equal to 1008 Hz

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Question 5 of 10
A dog pushes against the front door for 3 seconds with a force of 88 N. What
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A. 264 kg•m/s
O B. 77.2 kg•m/s
O C. 29.3 kg•m/s
OD. 128 kg•m/s
SLEMT

Answers

Answer:

265 Kg.m/s

Explanation:

Impulse = Force × time

So 88× 3 =264!!

Answer:

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An initially motionless test car is accelerated to 115 km/h in 8.58 s before striking a simulated deer. The car is in contact with the faux fawn for 0.815 s, after which the car is measured to be traveling at 60.0 km/h. What is the magnitude of the acceleration of the care before the collision? What is the magnitude of the average acceleration of the car during the collision?

Answers

Answer:

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Explanation:

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as part of rest the v₀ = 0

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Let's reduce the magnitudes to the SI system

              v = 115 km / h (1000 m / 1km) (1h / 3600s)

              v = 31.94 m / s

              v₂ = 60 km / h = 16.66 m / s

l

et's calculate

             a = 31.94 / 8.58

             a = 3.72 m / s²

b) For the operational average during the collision let's use the relationship between momentum and momentum

            I = Δp

            F Δt = m v_f - m v₀

            F = [tex]\frac{m ( v_f - v_o)}{t}[/tex]

            F = m [16.66 - 31.94] / 0.815

            F = m (-18.75)

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            -18.75 m = m a

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Answers

Answer:

125 joules

Explanation:

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yes

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Answers

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Explanation:

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ventral side:
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here are some pictures

Answers

Answer:

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Posterior end:C

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Answers

Answer:

125N

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Answers

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Answers

Answer:

Distance, S = 15m

Explanation:

Given the following data;

Mass = 4kg

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Initial velocity = 5m/s

Final velocity = 10m/s

To find the distance;

First of all, we would calculate the acceleration

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10 = 4*acceleration

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Acceleration = 2.5m/s²

Now to find the distance, we would use the third equation of motion.

[tex] V^{2} = U^{2} + 2aS [/tex]

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V represents the final velocity measured in meter per seconds. U represents the initial velocity measured in meter per seconds. a represents acceleration measured in meters per seconds square.

S represents the displacement measured in meters.

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10² = 5² + 2*2.5*S

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Answers

Answer:

Temperature or thermal energy.

Explanation:

Conduction involves the transfer of electric charge or thermal energy due to the movement of particles. When the conduction relates to electric charge, it is known as electrical conduction while when it relates to thermal energy, it is known as heat conduction.

In the process of heat conduction, thermal energy is usually transferred from fast moving particles to slow moving particles during the collision of these particles. Also, thermal energy is typically transferred between objects that has different degrees of temperature and materials (particles) that are directly in contact with each other but differ in their ability to accept or give up electrons.

Hence, the temperature or thermal energy of matter depends on how much the particles are moving, which depends on the amount of kinetic energy the particles possess.​

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Answers

Answer:

chemical energy

Explanation:

Gives you energy to burn off

Answer:

yea what she said

Explanation:

what ever she said

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Answers

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Answers

Answer:

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Explanation:

Yes; it is a solid

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Answers

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