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
<h3>Solution for the above question : -</h3>
Ohm's law states that :
the terms used are :
let's solve for electric current :

Answer:
V = 3.17 m/s
Explanation:
Given
Mass of the professor m = 85.0 kg
Angle of the ramp θ = 30.0°
Length travelled L = 2.50 m
Force applied F = 600 N
Initial Speed u = 2.00 m/s
Solution
Work = Change in kinetic energy

Weight = (mass) x (gravity)
Weight = (7.0 kg) x (gravity)
On Earth, where (gravity) is roughly 10 N/kg . . .
Weight = (7.0 kg) x (roughly 10 N/kg)
Weight = roughly 70 Newtons
That's <em>B </em>on Earth.
It would be some other number on other bodies.