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marusya05 [52]
3 years ago
9

After a plant or animal dies the 14C content decreases with a half-life of 5730 years. If an archaeologist finds an ancient fire

pit containing partially consumed firewood, and the 14C content of the wood is only 15.5% that of an equal carbon sample from a present-day tree, what is the age (in years) of the ancient site?
Physics
1 answer:
Shalnov [3]3 years ago
6 0

Answer:

age of the site is 15411.75 years old

Explanation:

Given data

plant or animal dies = 14C

time period = 5730 year

carbon = 15.5%

to find out

age (in years) of the ancient site

solution

we know that Final value = Initial value  × 0.5^{n}

here n is half life passed

so for 15.5%

15.5% = 100% of  0.5^{n}

0.155 = 1 × 0.5^{n}

now take log both side

log 0.155 = log  0.5^{n}

n = log 0.155 / log 0.5

n = 2.68966

we know here  5730 years in half life

so for 2.68966 half-lives = 2.68966 × 5730  =  15411.7518

age of the site is 15411.75 years old

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anygoal [31]
Aristotle created and it’s credited as the creator.
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3 years ago
You add 500 mL of water at 10°C to 100 mL of water at 70°C. What is the
Sphinxa [80]

Answer:

Option (c) : 20°C

Explanation:

t(final) =  \frac{w1 \times t1 + w2 \times t2}{w1 + w2}

T(final) = 500* 10 + 100*70/600 = 20°C

4 0
3 years ago
Due to the wave nature of light, light shined on a single slit will produce a diffraction pattern? Green light (520 nm) is shine
TiliK225 [7]

Answer:

Yes, it will produce a diffraction pattern.

a. 3.9 mm b. 1.95 mm

Explanation:

The light shined from a single slit will produce a diffraction pattern because,  the wavefront act as wavelets which generates its own wave according to Huygens principle. This therefore causes the diffraction pattern.

Given

wavelength of green light, λ = 520 nm = 520 × 10⁻⁹ m = 5.20 × 10⁻⁷ m

width of slit, d = 0.440 mm = 0.44 × 10⁻³ m = 4.4 × 10⁻⁴ m

Distance of slit from central maximum , D = 1.65 m

Distance of first minimum from central maximum, y = ?

a. The relationship between the slit width and wavelength is given by [tex} dsinθ = mλ [/tex]where d = slit width, θ = angular distance from central maximum, λ = wavelength of light and m = ±1, ±2, ±3...

The relationship between y and D is given by tanθ = y/D

Since θ is small, sinθ ≈ θ ≈ tanθ

so, dθ = mλ ⇒ θ = mλ/d = y/D

Therefore, y = mλD/d

Now, for the first minimum above the slit, m = +1 and for the first minimum below the slit, m = -1. So, y₁ =  λD/d and y₋₁ =  -λD/d. So, the width of the central maximum Δy is the difference between the first minima below and above the central maximum. So, Δy = y₁ - y₋₁ = λD/d -(-λD/d) = 2λD/d

Substituting the values from above, Δy= 2 × 5.20 × 10⁻⁷ × 1.65/4.4 × 10⁻⁴ =  3900 × 10⁻⁶ m = 3.9 × 10⁻³ m = 3.9 mm

b. The first order fringe is the fringe located between the first minimum and the second minimum. From dsinθ = mλ and tanθ = y/D when θ is small, sinθ ≈ θ ≈ tanθ. So, y = mλD/d. Let m= 1 and m=2 be the first and second minima respectively. So,y₁ =  λD/d and y₂ =  2λD/d. The difference Δy₁ = y₂ - y₁ is the width of the first order fringe. Therefore, Δy₁ = 2λD/d - λD/d= λD/d. Substituting the values from above, we have

λD/d= 5.20 × 10⁻⁷ × 1.65/4.4 × 10⁻⁴= 1.95 × 10⁻³ m = 1.95 mm

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3 years ago
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3 years ago
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9.
Ghella [55]

Given parameters:

Mass of the body  = 200g

Force on the body  = 10N

Unknown parameters:

Acceleration produced by the force  = ?

To solve this problem we must first define force in terms of mass and acceleration. This is possible due to the Newton's first law of motion.

  Force  = mass x acceleration

Here the unknown is acceleration and we can easily solve for it.

But we must take the mass to kilogram in order for it to cancel out.

        1000g  = 1 kg

        200g  = x kg =   \frac{200}{1000}   = 0.2kg

Now input the parameters and solve;

         10  = 0.2 x acceleration

   Acceleration  = \frac{10}{0.2}   = 50m/s²

The acceleration produced by the body is 50m/s²

4 0
3 years ago
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