Answer:
Well, to be perfectly honest in my humble opinion, of course without offending anyone who thinks different from my point of view but also by looking into this matter in a different perspective and without condemning one's view and by trying to make it objectified and considering each and everyone's valid opinion, I honestly believe that I completely forgot what I was going to say.
Answer:
Ki = 0.28665 J
h = 0.133 m
Explanation:
Given:
- The mass of rod M = 0.22 kg
- The length of rod L = 1.2 m
- The angular speed at the lowest point w = 2.33 rad /s
Find:
(a) the rod's kinetic energy at its lowest position
(b) how far above that position the center of mass rises.
Solution:
- The moment of inertia of a rod pivoted at one of its ends is given by I:
I = ML^2 / 3
- The Kinetic energy at the lowest point is given by the rotational energy as follows:
Ki = 0.5*I*w^2
Ki = 0.5*ML^2 / 3*w^2
Ki = ML^2*w^2 / 6
Ki = (.22*1.2^2*2.33^2) / 6
Ki = 0.28665 J
- Since no external force was acting on the rod we can apply the conservation of energy of system consisting of the rod where the change in kinetic energy leads to a change in gravitational potential Energy:
Kf - Ki = Pf - Pi
0 - 0.28665 J = mg( 0 - h )
mg*h = 0.28665
h = 0.28665 / ( 0.22*9.81 )
h = 0.133 m
Rate of speed (3 m/s north is three miles per second north, so it's a rate of speed)
Answer:
The horizontal distance of the target should be 2721,4 meters.
Explanation:
First of all we need to find the time that the emergency package hits the ground after the moment of release:
y=0 (because when it hits the ground it is on the level of 0m);
The emergency package hits the ground after 24,74 seconds from release.
Lets assume that package preserves his 110 m/s horizontal speed during the free fall. The targets horizontal distance is:
2721,4 meters
A)acceleration is in the direction of motion