6 . . . . . a crest
7 . . . . . the amplitude
8 . . . . . the wavelength
9 . . . . . a trough
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:
Well, newer telephone circuits built during the last decade are based on the digital transmission, not on the analog transmission. So it's the digital transmission circuit that has made the higher quality. Digital circuits converts the voice signals into the binary codes which is then translated again into the voice signal at the receiving end.
The answer is false.
Explanation:
Answer:
a) 4.9*10^-6
b) 5.71*10^-15
Explanation:
Given
current, I = 3.8*10^-10A
Diameter, D = 2.5mm
n = 8.49*10^28
The equation for current density and speed drift is
J = I/A = (ne) Vd
A = πD²/4
A = π*0.0025²/4
A = π*6.25*10^-6/4
A = 4.9*10^-6
Now,
J = I/A
J = 3.8*10^-10/4.9*10^-6
J = 7.76*10^-5
Electron drift speed is
J = (ne) Vd
Vd = J/(ne)
Vd = 7.76*10^-5/(8.49*10^28)*(1.60*10^-19)
Vd = 7.76*10^-5/1.3584*10^10
Vd = 5.71*10^-15
Therefore, the current density and speed drift are 4.9*10^-6
And 5.71*10^-15 respectively
Answer;
It does not have an hypothesis
Explanation;
The goal of descriptive investigation is to describe. It should provide factual, accurate and systematic descriptions of phenomena without attempting to infer causal relationships. It does not answer questions about the how, when, or why a particular phenomenon occurred.
A descriptive investigation includes a question but no hypothesis. Observations are recorded, but no comparisons are made and no variables are manipulated.
