Yo sup??
since the collision is elastic therefore we can say that the two balls will then move in opposite direction.
If ball 1 was moving from east to west then after collision it will move from west to east
and if ball 2 was moving from west to east then it will start moving from east to west.
Hope this helps.
I don't think that 4m has anything to do with the problem.
anyway. here.
A___________________B_______C
where A is the point that the train was released.
B is where the wheel started to stick
C is where it stopped
From A to B, v=2.5m/s, it takes 2s to go A to B so t=2
AB= v*t = 2.5 * 2 = 5m
The train comes to a stop 7.7 m from the point at which it was released so AC=7.7m
then BC= AC-AB = 7.7-5 = 2.7m
now consider BC
v^2=u^2+2as
where u is initial speed, in this case is 2.5m/s
v is final speed, train stop at C so final speed=0, so v=0
a is acceleration
s is displacement, which is BC=2.7m
substitute all the number into equation, we have
0^2 = 2.5^2 + 2*a*2.7
0 = 6.25 + 5.4a
a = -6.25/5.4 = -1.157
so acceleration is -1.157m/(s^2)
The velocity at the maximum height will always be 0. Therefore, you will count your final velocity as 0, and your initial velocity as 35 m/s. Next, we know that the acceleration will be 9.8 m/s^2. How? Because the ball is thrown directly upward, and the only force acting on it will be the force of gravity pushing it back down.
The formula we use is h = (Vf^2 - Vi^2) / (2*-9.8m/s^2)
Plugging everything in, we have h = (0-1225)/(19.6) = 62.5 meters is the maximum height.
Answer:
one im so sry i have no idea. I have been researshing for about 30min and i cant find anything im so sry:/
Explanation:
Answer:
2.75 m/s^2
Explanation:
The airplane's acceleration on the runway was 2.75 m/s^2
We can find the acceleration by using the equation: a = (v-u)/t
where a is acceleration, v is final velocity, u is initial velocity, and t is time.
In this case, v is 71 m/s, u is 0 m/s, and t is 26.1 s Therefore: a = (71-0)/26.1
a = 2.75 m/s^2
