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
Answer:
0.8726 
Explanation:
We are to convert 1.85 x 
to
First, let us convert the numerator from ft3 to m3
1 ft3 = 0.0283 m3
Hence,
1.85 x ft3 = 1.85 x x 0.0283 m3
= 52.355 m3
Now, let us convert the denominator from minutes to seconds
1 min = 60 sec
Therefore;
1.85 x = 52.355/60
= 0.8726
Answer:
1/8
Explanation:
17,100 years is 3 times the half-life of 5,700 years. After each half-life, half remains, so the amount remaining after 3 half-lives is ...
(1/2)(1/2)(1/2) = 1/8
1/8 of the sample remains after 17,100 years.
The appropriate response is Zero degrees. The beam will leave the two mirrors along a way parallel to the one it came in on. This is the guideline of the corner reflector, which is frequently utilized as a radar target. Take note of that the corner reflector utilizes three reflecting surfaces (that are set up at 90o from each other) rather than the two like are being utilized here. Wikipedia has a truly awesome drawing that shows this two-dimentional issue pleasantly. A moment connection is given to the article on the corner reflector and the 3-D angles.
