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3 years ago

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The ratio of carbon-14 to nitrogen-14 is an artifact is 1:3. Given that half-life of carbon-14 is 5730years, how old is the arti

fact?

1 answer:

dusya [7]3 years ago

5 0

Answer:

9155 years old

Explanation:

We use the following expression for the decay of a substance:

$N = N_0\,\,e^{-k*t}$

So we first estimate the value of k knowing that the half-life of the C14 is 5730 years:

$N = N_0\,\,e^{-k*t}\\N_0/2=N_0\,\,e^{-k*5730}\\1/2 = e^{-k*5730}\\ln(1/2)=-k*5730\\k= 0.00012$

so, now we can estimate the age of the artifact by solving for"t" in the equation:

$1/3=e^{-0.00012*t}\\ln(1/3)= -0.00012*t\\t=9155. 102$

which we can round to 9155 years old.

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Answer:

Answer:

Explanation:

Given that

K=8.98755×10^9Nm²/C²

Q=0.00011C

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g=9.8m/s²

1. The electric field inside a conductor is zero

εΦ=qenc

εEA=qenc

net charge qenc is the algebraic sum of all the enclosed positive and negative charges, and it can be positive, negative, or zero

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2. Since the entire charge us inside the surface, then the electric field at a distance r (5.2m) away form the surface is given as

F=kq1/r²

F=kQ/r²

F=8.98755E9×0.00011/5.2²

F=36561.78N/C

The electric field at the surface of the conductor is 36561N/C

Since the charge is positive the it is outward field

3. Given that a test charge is at 12.6m away,

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E=kQ/r²

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3 years ago

Evaluate the expression a+b when a=23 and b=45 . Write in simplest form

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If a = 23 and b = 45

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2 years ago

Two objects are moving in the xy-plane. Object A has a mass of 3.2 kg and has a velocity = (2.3 m/s)i+ (4.2 m/s)j and object B h

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The total momentum of the system is 2.14i + 21.27j.

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The total momentum of the system is defined as follows:

As momentum is vector quantity and vectors can be added, so, the momentum of a system is given by

P = Pₓ + P'

where Pₓ is the x-component of momentum

P' is the y-component of the momentum

Also, we know that

P=mv

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v is velocity

Thus,

P = Pₓ + P'

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v' is the y- component of the velocity

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Now,

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1 year ago

A tugboat tows a ship with a constant force of magnitude F1. The increase in the ship's speed during a 10 s interval is 3.0 km/h

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Answer:

The magnitude of F₁ is 3.7 times of F₂

Explanation:

Given that,

Time = 10 sec

Speed = 3.0 km/h

Speed of second tugboat = 11 km/h

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$v_{1}=\dfrac{3.0\times10^{3}}{3600}$

$v_{1}=0.833\ m/s$

The force F₁is constant acceleration is also a constant.

$F_{1}=ma_{1}$

We need to calculate the acceleration

Using formula of acceleration

$a_{1}=\dfrac{v}{t}$

$a_{1}=\dfrac{0.833}{10}$

$a_{1}=0.083\ m/s^2$

Similarly,

$F_{2}=ma_{2}$

For total force,

$F_{3}=F_{2}+F_{1}$

$ma_{3}=ma_{2}+ma_{1}$

The speed of second tugboat is

$v=\dfrac{11\times10^{3}}{3600}$

$v=3.05\ m/s$

We need to calculate total acceleration

$a_{3}=\dfrac{v}{t}$

$a_{3}=\dfrac{3.05}{10}$

$a_{3}=0.305\ m/s^2$

We need to calculate the acceleration a₂

$0.305=a_{2}+0.083$

$a_{2}=0.305-0.083$

$a_{2}=0.222\ m/s^2$

We need to calculate the factor of F₁ and F₂

Dividing force F₁ by F₂

$\dfrac{F_{1}}{F_{2}}=\dfrac{m\times0.83}{m\times0.22}$

$\dfrac{F_{1}}{F_{2}}=3.7$

$F_{1}=3.7F_{2}$

Hence, The magnitude of F₁ is 3.7 times of F₂

3 0

2 years ago