Answer:
A. It does not exhibit projectile motion and follows a straight path down the ramp.
Vf^2 = Vi^2 + 2ad
a= 34 m/s^2
Vi = 0 m/s
d = 3400m
Vf = 480.83 m/s
a=v/t
t=v/a
t=480.83/34
t=14.142 s
Answer:
<em>1,378.9ms²</em>
Explanation:
Given the following
Distance S = 70.6m
Time t = 0.32secs
Initial velocity = 0m/s
Required
Acceleration
Using the equation of motion
S = ut+1/2at²
Substitute
70.6 = 0+1/2a(0.32)²
70.6 = 0.0512a
a = 70.6/0.0512
a = 1,378.9
<em>Hence the acceleration is 1,378.9ms²</em>
Since momentum is a vector quantity, take any direction as positive and other as negative. Answer won't change.
Answer:
120 J
Explanation:
KE = mv²/2 = (0.15 kg * [40 m/s]²)/2 = 120 J
