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Nat2105 [25]
3 years ago
12

The theory of nuclear astrophysics is that all the heavy elements like uranium are formed in the interior of massive stars. Thes

e stars eventually explode, releasing these elements into space. If we assume that at the time of the explosion there were equal amount of U-235 and U-238, how long ago did the star(s) explode that released the elements that formed our Earth? The present U-235/U-238 ratio is 0.00700. [The half-lives of U-235 and U-238 are 0.700 × 109 yr and 4.47 × 109 yr.]
Physics
1 answer:
joja [24]3 years ago
8 0

Answer:

t = 5.94x10⁹ years.

Explanation:

The time of the explosion can be calculated using the decay equation:

N_{t} = N_{0}e^{-\lambda t}

<u>Where:</u>

N(t): is the quantity of the element at the present time

N(0): is the quantity of the element at the time of explosion

λ: is the decay constant

t: is the time

Knowing that the present U-235/U-238 ratio is 0.00700 and that at the time of the explosion there were equal amount of U-235 and U-238, we have:

\frac{N_{U-235}}{N_{U-238}} = \frac{N_{U-235_{0}}e^{-\lambda_{U-235} t}}{N_{U-238_{0}}e^{-\lambda_{U-238} t}}     (1)

The decay constant is equal to:

\lambda = \frac{ln(2)}{t_{1/2}}  

For the U-235 we have:

\lambda_{U-235} = \frac{ln(2)}{0.700 \cdot 10^{9} y} = 9.90 \cdot 10^{-10} y^{-1}

For the U-238 we have:

\lambda_{U-238} = \frac{ln(2)}{4.47 \cdot 10^{9} y} = 1.55 \cdot 10^{-10} y^{-1}

By introducing the values of \lambda_{U-235} and \lambda_{U-238} into equation (1) we have:

0.00700 = \frac{e^{-9.90 \cdot 10^{-10} t}}{e^{-1.55 \cdot 10^{-10} t}}        

0.00700 = e^{(-9.90 \cdot 10^{-10} + 1.55 \cdot 10^{-10}) t}    

ln(0.00700) = (-9.90 \cdot 10^{-10} + 1.55 \cdot 10^{-10}) t            

t = \frac{ln(0.00700)}{-9.90 \cdot 10^{-10} + 1.55 \cdot 10^{-10}} = 5.94 \cdot 10^{9} y

Therefore, the star exploded 5.94x10⁹ years ago.

I hope it helps you!    

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A satellite is in a circular orbit 21000 km above the Earth’s surface; i.e., it moves on a circular path under the influence of
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Answer:

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Explanation:

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radius of the orbit (r) = 21000 km + 6.37 × 10⁶ m

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                                  = 2.737 × 10⁷m

Orbital Speed (V₀) = √(r × g)

Orbital Speed (V₀) = √(2.737 × 10⁷  × 0.532655 )

                              = 3818.215

                              = 3.818 × 10³

                             = 3.818 Km

To find the time it takes to complete one orbit around the Earth; we use the formula:

Time (t) = 2 π × \frac{r}{V_o}

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            = 4.5 × 10 ⁴ s

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