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
It is about 100oC at a pressure of 1.1 atmosphere. Hope this helps.
<u>We are Given:</u>
Mass of the block (m) = 500 grams or 0.5 Kg
Initial velocity of the block (u) = 0 m/s
Distance travelled by the block (s) = 8 m
Time taken to cover 8 m (t)= 2 seconds
Acceleration of the block (a) = a m/s²
<u>Solving for the acceleration:</u>
From the seconds equation of motion:
s = ut + 1/2* (at²)
<em>replacing the variables</em>
8 = (0)(2) + 1/2(a)(2)²
8 = 2a
a = 4 m/s²
Therefore, the acceleration of the block is 4 m/s²
It's 60.... i have used the formula
L=r^2mw
and didn't use the value 0.75...
During the fall, all the initial potential energy of the rock

has converted into kinetic energy of motion

where h is the initial height of the rock, m its mass, and v its velocity just before hitting the water. So, for energy conservation, we have

and so we can find the value of K, the kinetic energy of the rock just before hitting the ground: