Answer:
we assume that it starts with a velocity of 10m/s. At 2m height above ground level, its velocity decreases at 3m above ground level
for its way down the velocity at 3m on its way down is more than its velocity at 2m on its way down.
Explanation:
A student throws a small rock straight upwards. The rock rises to its highest point and then falls back down. How does the speed of the rock at 2m on the way down compare with its speed at 2m on the way up?
It decreases in speed on its way down and increases in speed on its way down.
it decreases in speed on its way up because the the vertical motion is against the earths gravitational pull on an object to the earth's center
.It increases in speed on his way down because its under the influence of gravity
from newton's equation of motion we can check by
using V^2=u^2+2as
we assume that it starts with a velocity of 10m/s. At 2m height above ground level, its velocity decreases at 3m above ground level
for its way down the velocity at 3m on its way down is more than its velocity at 2m on its way down.
final velocity = initial
velocity + (acceleration x time) <span>
3.9 m/s = 0 m/s + (acceleration x 0.11 s)
3.9 m/s / 0.11 s = acceleration
30.45 m/s^2 = acceleration
distance = (initial velocity x time) +
1/2(acceleration)(time^2)
distance (0 m/s x 0.11 s) + 1/2(30.45 m/s^2)(0.11s ^2)
<span>distance = 0.18 m</span></span>
Answer:
I would say the net force acting on the car is in the opposite direction of the car's motion is correct
The answer is B.
This is because you add up all of the times (1.44s+1.70s+1.58s+1.76s) and you get 6.48 then you divide 6.48 by 4 to get the average of the times. Now you get the distance (200m) and because speed=distance/time you divide 200m/1.62s to get 123m/s. I hope this made sense :)