(a) Assuming that Q satisfies the differential equation Q' = -rQ, determine the decay constant r for carbon-14. (b) Find an expression for Q(t) at any time t, if Q(0) = Qo. (c) Suppose that certain remains are discovered in which the current residual amount of carbon-14 is 20% of the original amount. Determine the age of these remains.
Answer:
a) r = (In 2)/(t1/2) = (In 2)/5730 = 0.000121/year
b) Q(t) = Q₀ (e^-rt)
c) Are of the 20% remnant of Carbon-14 = 13301.14 years.
Step-by-step explanation:
Q' = -rQ
Q' = dQ/dt
dQ/dt = -rQ
dQ/Q = -rdt
Integrating the left hand side from Q₀ to Q₀/2 and the right hand side from 0 to t1/2 (half life, t1/2 = 5730 years)
In ((Q₀/2)/Q₀) = -r(t1/2)
In (1/2) = -r(t1/2)
In 2 = r(t1/2)
r = (In 2)/(t1/2) = (In 2)/5730 = 0.000121 /year
b) Q' = -rQ
Q' = dQ/dt
dQ/dt = -rQ
dQ/Q = -rdt
Integrating the left hand side from Q₀ to Q(t) and the right hand side from 0 to t.
In (Q(t)/Q₀) = -rt
Q(t)/Q₀ = e^(-rt)
Q(t) = Q₀ (e^-rt)
c) Q(t) = Q₀ (e^-rt)
Q(t) = 0.2Q₀, t = ? and r = 0.000121/year
0.2Q₀ = Q₀ (e^-rt)
0.2 = e^-rt
In 0.2 = -rt
-1.6094 = - 0.000121 × t
t = 1.6094/0.000121 = 13301.14 years.
Hope this Helps!
A. the position of the figure changes
The answer to the question is y=2
Mpg = 476/14 = 34 mpg
d = 578 miles = (34 mpg)(x gallons)
578 = 34x
x = 578/34 = 17 gallons
