Answer:
All these statement are true
Explanation:
Gravity will be acting like a centripetal force for the circular motion of object around earth, which makes it perpendicular to the velocity vector. In the case of elliptical motion, gravity can still be divided into 2 vectors, one parallel and the other perpendicular to the velocity. At the nearest point in elliptical motion, gravity is directly perpendicular to velocity just like in circular motion. At the farthest point, the potential energy is minimized and has been converted into kinetic energy. Therefore at this point the speed is greatest.
The guy below is wrong!
F=ma
Where force = mass x acceleration
We dont have acceleration, a= change in velocity divided by the time taken.
a = v (final velocity) - u (initial) / t
a us 8-0 (at rest means u was 0) / 20 = 0.4
Using F=ma
F= mass x acceleration
F= 4 x 0.4
F=1.6 N
Answer:
x = 7.62 m
Explanation:
First we need to calculate the weight of the rocket:
W = mg
we will use the gravity as 9.8 m/s². We have the mass (500 g or 0.5 kg) so the weight is:
W = 0.5 * 9.8 = 4.9 N
We know that the rocket exerts a force of 8 N. And from that force, we also know that the Weight is exerting a force of 4.9. From here, we can calculate the acceleration of the rocket:
F - W = m*a
a = F - W/m
Solving for a:
a = (8 - 4.9) / 0.5
a = 6.2 m/s²
As the rocket is accelerating in an upward direction, we can calculate the distance it reached, assuming that the innitial speed of the rocket is 0. so, using the following expression we will calculate the time which the rocket took to blast off:
y = vo*t + 1/2 at²
y = 1/2at²
Solving for t:
t = √2y/a
t = √2 * 20 / 6.2
t = √6.45 = 2.54 s
Now that we have the time, we can calculate the horizontal distance:
x = V*t
Solving for x:
x = 3 * 2.54 = 7.62 m
Answer:
(1) 2.25m/s^2
(2) 45.6m
Explanation:
(1) A car accelerates uniformly from 12m/s to 39m/s in 12 seconds
Therefore the average acceleration can be calculated as follows
a = 39-12/12
a = 27/12
a= 2.25m/s^2
(2) A butterfly is flying at 4m/s , it accelerates uniformly at 1.2 m/s for 6 seconds
u= 4
a= 1.2
t= 6
Therefore the distance can be calculated as follows
S= ut + 1/2at^2
= 4×6 + 1/2 × 1.2 × 6^2
= 24 + 1/2 × 1.2 × 36
= 24 + 1/2 × 43.2
= 24 + 21.6
S = 45.6m
Hence the butterfly travels at 45.6m
