An object with non-zero mass (even negligible mass is non-zero) will never reach the speed of light. Due to relativistic effects, each "unit" of acceleration becomes less effective at increasing your velocity (relative to some other object, of course) as your relative velocity approaches the speed of light.
And even if there was a way, If you would accelerate to the 99,99% of the speed light in just 1 second, you would experience a G-force of aprox. 30,600,000 g's which is enough to kill you in a few seconds
Answer : The value of the constant for a second order reaction is, 
Explanation :
The expression used for second order kinetics is:
![kt=\frac{1}{[A_t]}-\frac{1}{[A_o]}](https://tex.z-dn.net/?f=kt%3D%5Cfrac%7B1%7D%7B%5BA_t%5D%7D-%5Cfrac%7B1%7D%7B%5BA_o%5D%7D)
where,
k = rate constant = ?
t = time = 17s
![[A_t]](https://tex.z-dn.net/?f=%5BA_t%5D)
= final concentration = 0.0981 M
![[A_o]](https://tex.z-dn.net/?f=%5BA_o%5D)
= initial concentration = 0.657 M
Now put all the given values in the above expression, we get:
Therefore, the value of the constant for a second order reaction is,
The shape is connected in parallel so;
5.1) Ans;
5.2) Ans;
I hope I helped you^_^
Answer:
is it bad if i keep thinking about p ussy
Explanation:
a) we can answer the first part of this by recognizing the player rises 0.76m, reaches the apex of motion, and then falls back to the ground we can ask how
long it takes to fall 0.13 m from rest: dist = 1/2 gt^2 or t=sqrt[2d/g] t=0.175
s this is the time to fall from the top; it would take the same time to travel
upward the final 0.13 m, so the total time spent in the upper 0.15 m is 2x0.175
= 0.35s
b) there are a couple of ways of finding thetime it takes to travel the bottom 0.13m first way: we can use d=1/2gt^2 twice
to solve this problem the time it takes to fall the final 0.13 m is: time it
takes to fall 0.76 m - time it takes to fall 0.63 m t = sqrt[2d/g] = 0.399 s to
fall 0.76 m, and this equation yields it takes 0.359 s to fall 0.63 m, so it
takes 0.04 s to fall the final 0.13 m. The total time spent in the lower 0.13 m
is then twice this, or 0.08s
