a) $\frac{dm}{dt} = -k\cdot m$ , b) $m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }$ , c) $m(t) = 10\cdot e^{-\frac{t}{2438.155} }$ , d) $m(300) \approx 8.842\,g$

Step-by-step explanation:

a) Let assume an initial mass m decaying at a constant rate k throughout time, the differential equation is:

$\frac{dm}{dt} = -k\cdot m$

b) The general solution is found after separating variables and integrating each sides:

$m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }$

Where $\tau$ is the time constant and $k = \frac{1}{\tau}$

c) The time constant is:

$\tau = \frac{1690\,yr}{\ln 2}$

$\tau = 2438.155\,yr$

The particular solution of the differential equation is:

$m(t) = 10\cdot e^{-\frac{t}{2438.155} }$

d) The amount of radium after 300 years is:

$m(300) \approx 8.842\,g$

4 0

3 years ago

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Look at the pic and help if you can thxxxx

givi [52]

Answer:

The 1,3,and 4.

Step-by-step explanation:

6 0

3 years ago