Answer:
8. 2.75·10^-4 s^-1
9. No, too much of the carbon-14 would have decayed for radiation to be detected.
Explanation:
8. The half-life of 42 minutes is 2520 seconds, so you have ...
1/2 = e^(-λt) = e^(-(2520 s)λ)
ln(1/2) = -(2520 s)λ
-ln(1/2)/(2520 s) = λ ≈ 2.75×10^-4 s^-1
___
9. Reference material on carbon-14 dating suggests the method is not useful for time periods greater than about 50,000 years. The half-life of C-14 is about 5730 years, so at 65 million years, about ...
6.5·10^7/5.73·10^3 ≈ 11344
half-lives will have passed. Whatever carbon 14 may have existed at the time will have decayed completely to nothing after that many half-lives.
Answer:
El área de la placa es aproximadamente 5102.752 centímetros cuadrados.
Explanation:
Asumamos que el cambio dimensional como consecuencia de la temperatura es pequeña, entonces podemos estimar el área de la placa de cobre en función de la temperatura mediante la siguiente aproximación:
(1)
Donde:
- Ancho de la placa, en centímetros.
- Longitud de la placa, en centímetros.
- Coeficiente de dilatación, en
.
- Temperatura inicial, en grados Celsius.
- Temperatura final, en grados Celsius.
Si sabemos que
,
,
,
and
, entonces el área de la placa a la temperatura final:
![A_{f} = (65\,cm)\cdot (78\,cm)\cdot \left[1+\left(17\times 10^{-6}\,\frac{1}{^{\circ}C} \right)\cdot (400\,^{\circ}C-20\,^{\circ}C)\right]](https://tex.z-dn.net/?f=A_%7Bf%7D%20%3D%20%2865%5C%2Ccm%29%5Ccdot%20%2878%5C%2Ccm%29%5Ccdot%20%5Cleft%5B1%2B%5Cleft%2817%5Ctimes%2010%5E%7B-6%7D%5C%2C%5Cfrac%7B1%7D%7B%5E%7B%5Ccirc%7DC%7D%20%5Cright%29%5Ccdot%20%28400%5C%2C%5E%7B%5Ccirc%7DC-20%5C%2C%5E%7B%5Ccirc%7DC%29%5Cright%5D)

El área de la placa es aproximadamente 5102.752 centímetros cuadrados.
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Answer:
In a chemical reaction, the atoms and molecules that interact with each other are called reactants. ... No new atoms are created, and no atoms are destroyed. In a chemical reaction, reactants contact each other, bonds between atoms in the reactants are broken, and atoms rearrange and form new bonds to make the products.
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Inversion will be your answer