When the force of the opponent's punch is extended, with time, the effect of the blow or the force of the blow is reduced thereby reducing chances for a knockout punch from the opponent to the boxer.
The 1st one goes two added sodoes the second one then the third goes to removed then the fourth goes to added and the rest go to removed
Using the equation v(avg)=distance/time
and the equation v=v(original)+a(t)
solve for acceleration
2600=0+a(12)
a=216.66666 m/s^2
Then, you use the equation
v^2=v(original)+2a*(change in x)
2600^2=2(216.666666)*change in x
6760000/2/216.666666 = 15600 meters which is the length of the race
Then using v(avg)=x/t
15600/12= 1300 m/s
For the first part, we are looking for Vf when dy=11.0
Upward is positive, downward is negative.
So <span>Vf = square root [2(-9.8)(11.0) + (18.0)^2] </span>
<span>Vf = 10.4 m/s your answer is correct.
For the part b, t is equals to the time took to reach and dy is equals to 11.0
you did, </span>11= 18t m/s-(1/2) 9.8t^2 then <span>-11 + 18t- 9.8t^2. By quadratic formula, for the way down the answer is 2.9 s while on it's way up, the answer is 0.77 s</span><span>
</span>
Answer:
Explanation:
we know that
as we see that
relative error
Where X_1 IS HEIGHT OF ROCK
IS THE HEIGHT OF ROAD
= uncertainity in measuring distance
Putting all value to get uncertainity in angle
solving for we get