Answer:
B AND D
HOPE ITS HELF FUL .........
The car travels a distance <em>d</em> from rest with acceleration <em>a</em> after time <em>t</em> of
<em>d</em> = 1/2 <em>a</em> <em>t</em>²
It covers 69 m with 2.8 m/s² acceleration, so that
69 m = 1/2 (2.8 m/s²) <em>t</em>²
<em>t</em>² = 2 (69 m) / (2.8 m/s²)
<em>t</em> ≈ 7.02 s
where we take the positive square root because we're talking about time *after* the car begins accelerating.
Answer:
Change in momentum, 
Explanation:
It is given that,
Mass of the basketball, m = 601 g = 0.601 kg
The basketball makes an angle of 29 degrees to the vertical, it hits the floor with a speed, v = 6 m/s
It bounces up with the same speed, again moving to the right at an angle of 29 degree to the vertical. We need to find the change in momentum. It is given by :




So, the change in momentum of the basketball is 6.3 kg-m/s. Hence, this is the required solution.
Answer:
I am not really sure but i think its option 2
Explanation: