A scientist has 37 grams of a radioactive substance that decays exponentially. Assuming k=-0.3,how many grams of radioactive substance remain after 9 days.
how is this trig? Anyway, just plug in your numbers. Assuming a model of
A = Ao e^(kt), where t is in days, we have
37e^(-0.3*9) = ?
To find the amount of radioactive substance remaining after a certain time using exponential decay, we can use the formula:
A = A0 * e^(kt)
Where:
A is the amount of substance remaining after time t
A0 is the initial amount of substance
k is the decay constant
t is the time in question
In this case:
A0 = 37 grams (the initial amount)
k = -0.3 (the decay constant)
t = 9 days (the time in question)
Plugging these values into the formula, we get:
A = 37 * e^(-0.3 * 9)
Now let's calculate this step-by-step:
1. Calculate the product of the decay constant and time:
-0.3 * 9 = -2.7
2. Calculate the exponential value of -2.7 using Euler's number (e):
e^(-2.7) ≈ 0.067
3. Multiply the initial amount (37) by the exponential value:
37 * 0.067 ≈ 2.479 grams (rounded to three decimal places)
Therefore, approximately 2.479 grams of the radioactive substance remain after 9 days.
To determine how many grams of the radioactive substance remain after 9 days, we can use the exponential decay formula:
N(t) = N₀ * e^(kt)
Where:
N(t) = the amount of the substance remaining after time t
N₀ = the initial amount of the substance
e = the base of the natural logarithm (approximately 2.71828)
k = the decay constant
t = time (in this case, 9 days)
Given that k = -0.3 and N₀ = 37 grams, we can substitute these values into the formula to find N(9):
N(9) = 37 * e^(-0.3 * 9)
To calculate N(9), we need to use the value of e, which is approximately 2.71828. Plugging in these values, we get:
N(9) = 37 * 2.71828^(-0.3 * 9)
Using a calculator or computer program, we can calculate this value:
N(9) ≈ 37 * 2.71828^(-2.7)
N(9) ≈ 37 * 0.067
N(9) ≈ 2.479 grams
Therefore, after 9 days, approximately 2.479 grams of the radioactive substance remain.