Question

Alguém me ajuda, por favor.

Answer 1

$ Ola Lpl

M/M₀ = 1/2 = e^2k 

-ln(2) = 2k

k = -ln(2)/2 = -0.693/2 = -0.3465 

pronto 

Answer 2

$\text{Depois de dois anos, a massa inicial se divide pela metade.}\\\text{Portanto, temos que quando }t=2\text{, a massa }M=\dfrac{M_o}{2}.\\\\\text{Substituindo esses valores na fun\c{c}}\mathrm{\tilde{a}o,\ }\text{obtemos o valor do coeficiente de desintegra\c{c}}\mathrm{\tilde{a}o.}\\\\M=M_o\cdot\e^{kt}\\\\\\\left(\dfrac{M_o}{2}\right)=M_o\cdot{e}^{k(2)}\\\\\\\dfrac{M_o}{2}=M_o\cdot{e}^{2k}\\\\\\\dfrac{M_o}{2\cdot{M}_o}={e}^{2k}\\\\\\\dfrac{1}{2}={e}^{2k}\\\\2^{-1}=e^{2k}\\\\\ln(2^{-1})=\ln(e^{2k})\\\\-1\ln{2}=2k\\\\-1\cdot(0{,}693)=2k\\\\-0{,}693=2k\\\\k=-\dfrac{0{,}693}{2}\\\\\boxed{k=-0{,}3465}$