Answer:
t = 5.59x10⁴ y
Explanation:
To calculate the time for the ¹⁴C drops to 1.02 decays/h, we need to use the next equation:
(1)
<em>where 
: is the number of decays with time, A₀: is the initial activity, λ: is the decay constant and t: is the time.</em>
To find A₀ we can use the following equation:
(2)
<em>where N₀: is the initial number of particles of ¹⁴C in the 1.03g of the trees carbon </em>
From equation (2), the N₀ of the ¹⁴C in the trees carbon can be calculated as follows:
<em>where 
: is the tree's carbon mass, 
: is the Avogadro's number and 
: is the ¹²C mass. </em>
Similarly, from equation (2) λ is:
<em>where t 1/2: is the half-life of ¹⁴C= 5700 years </em>
So, the initial activity A₀ is:
Finally, we can calculate the time from equation (1):
I hope it helps you!
Answer:
The acceleration of man 1 and 2 is 
and 
.
Explanation:
Mass of man 1, m₁ = 80 kg
Mass of man 2, m₂ = 60 kg
One man pulls on the rope with a force of 250 N.
Let a₁ is acceleration of man 1,
F = m₁a₁
Let a₂ is acceleration of man 1,
F = m₂a₂
So, the acceleration of man 1 and 2 is and .
They are both in motion because an object is not at rest, but moving so slow it could be at rest. A car going at the same constant velocity is neither speeding up or slowing down, an object "at rest" is also moving at a constant rate, not speeding up or slowing done.
I think its B or D, most likely D.
