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:
a = w² r
Explanation:
In this exercise, indicate that the wheel has angular velocity w, the worm experiences the same angular velocity if it does not move, and has an acceleration towards the center of the circle, according to Newton's second law, called the centripetal acceleration.
a = v² / r
angular and linear variables are related
v = w r
we substitute
a = w² r
where r is the radius of the wheel
An ion is created by the transfer of electrons. The metals give away the elections and become positively charged. The non - metals take on electrons.
Balance.
So an ion is any atom that either gives away or takes on electrons.
Answer:
308,000 or 30.8×10^3
Explanation:
v=f×lamda
v is ?, f is 875Hz, lamda is 352m
v=875×352
v=308,000
v=30.8×10^3 m/s
Answer:
Explanation:
Given that
From the diagram
By differentiating with time t
When x= 10 m
θ = 64.53°
Now by putting the value in equation
Therefore rate of change in the angle is 0.038\ rad/s
