Answer:
The acceleration required by the rocket in order to have a zero speed on touchdown is 19.96m/s²
The rocket's motion for analysis sake is divided into two phases.
Phase 1: the free fall motion of the rocket from the height 2.59*102m to a height 86.9m
Phase 2: the motion of the rocket due to the acceleration of the rocket also from the height 86.9m to the point of touchdown y = 0m.
Explanation:
The initial velocity of the rocket is 0m/s when it started falling from rest under free fall. g = 9.8m/s² t1 is the time taken for phase 1 and t2 is the time taken for phase2.
The final velocity under free fall becomes the initial velocity for the accelerated motion of the rocket in phase 2 and the final velocity or speed in phase 2 is equal to zero.
The detailed step by step solution to the problems can be found in the attachment below.
Thank you and I hope this solution is helpful to you. Good luck.
The speed change : Δv = 0.41 m/s
<h3>Further explanation</h3>
Given
mass = 5.5 kg
Force = 15 N
time = 0.15 s
Required
the speed change
Solution
Newton 2nd's law
Impulse and momentum
F = m.a
F = m . Δv/t
F.t = m.Δv
Input the value :
15 N x 0.15 s = 5.5 kg x Δv
Δv = 0.41 m/s
You would have to place your sensor above earth's atmosphere because it blocks out nearly all x-rays. this is why we have the Chandra observatory
hope this helps
Answer:
0 - 60 mph = 0 - 26.8 m/s = 0 - 96.6 km/h; 0 - 100 km/h = 0 - 27.8 m/s = 0 - 62.1 mph.
Explanation:
Answer:17.08 s
Explanation:
Given
distance between First and second Runner is 45.6 m
speed of first runner=3.1 m/s
speed of second runner=4.65 m/s
Distance between first runner and finish line is 250 m
Second runner need to run a distance of 250+45.6=295.6 m
Time required by second runner
time required by first runner to reach finish line
Thus second runner reach the finish line 80.64-63.56=17.08 s earlier