The complete queston is The amount of a radioactive element A at time t is given by the formula
A(t) = A₀e^kt
Answer: A(t) =N e^( -1.2 X 10^-4t)
Explanation:
Given
Half life = 5730 years.
A(t) =A₀e ^kt
such that
A₀/ 2 =A₀e ^kt
Dividing both sides by A₀
1/2 = e ^kt
1/2 = e ^k(5730)
1/2 = e^5730K
In 1/2 = 5730K
k = 1n1/2 / 5730
k = 1n0.5 / 5730
K= -0.00012 = 1.2 X 10^-4
So that expressing N in terms of t, we have
A(t) =A₀e ^kt
A₀ = N
A(t) =N e^ -1.2 X 10^-4t
Solution= The answer is true
<span> Second-level consumer </span>
I believe that is true.
hope this helps!
I think you almost got it.
At the top, the velocity only has horizontal component, so v=12 m/s is v_x, which is v*cos(theta), because v_x is constant, so the same when it was launched or now.
With the value of the initial speed (28 m/s, which is the total speed), you can set
v_x = v * cos( theta ) ---> 12 = 28*cos(theta) --> cos(theta)=12/28=3/7
or theta = 64.62 deg, it is D. Think about it. I hope you see it.
