IM sure there is C, D, and E in kuiper belts, but not really sure of silicon and iron
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:
Acceleration = 10.06 m/s²
Explanation:
1 mile = 1.6093km
1609.3m = 1 mile
1 m = 
mile
50.0 miles/hour = 
m/s
= 22.35m/s
from equation
S = Ut + 1/2 at²
v = U + at
22.35 = 0 + a * 2.22
a = 22.35 ÷ 2.22
= 10.06 m/s²
It is difficult for astronomers to find object like planets and asteroids because it takes a lot of time to verify the objects locations and what surrounds a certain object in order to prove and be precise of its location
