Answer:
The required work done is 
Explanation:
Consider 'F' is the applied force on the crate and 'f' be the force created by friction. According to the figure if '
' be the coefficient of friction, then
where 'M', 'N' and 'g' are the mass of the crate, the normal force aced upon the block and the acceleration due to gravity respectively.
Since the application of force by the movers does not create any acceleration to the block, we can write
So the work done (W) in moving the crate by a distance s = 10.6 m is
Your average speed was
(100 m) / (13.8 s) = 7.25 m/s .
If you finished 0.001s ahead of him, then at your average speed, that corresponds to
(7.25 m/s) x (0.001 s) = 0.00725 m
That's 7.25 millimeters ... about 0.28 of an inch !
NOTE:. I think this is only valid if your speed was a constant ~7.25 m/s all the way.
If it does do it then yeah it will for it
Answer:
c) 2.02 x 10^16 nuclei
Explanation:
The isotope decay of an atom follows the equation:
ln[A] = -kt + ln[A]₀
<em>Where [A] is the amount of the isotope after time t, k is decay constant, [A]₀ is the initial amount of the isotope</em>
[A] = Our incognite
k is constant decay:
k = ln 2 / Half-life
k = ln 2 / 4.96 x 10^3 s
k = 1.40x10⁻⁴s⁻¹
t is time = 1.98 x 10^4 s
[A]₀ = 3.21 x 10^17 nuclei
ln[A] = -1.40x10⁻⁴s⁻¹*1.98 x 10^4 s + ln[3.21 x 10^17 nuclei]
ln[A] = 37.538
[A] = 2.01x10¹⁶ nuclei remain ≈
<h3>c) 2.02 x 10^16 nuclei</h3>
Answer:
anyone know this or should i get my brother
