That it's more puashe on the back of the canoe and that effects the back of the canoe to fall back
The answer to the question is an orgamisn
- keep you eyes on the ball, after you hit the ball you look up, you have to see the club hit the ball
- your grip it really matters on what your comfortable with but first you put your left hand on the club then put your right hand below your left hand, then put your right pinky on top of your index finger, but you can inter lock your pinky and index finger
- DON'T HIT AT THE BALL HIT THREW THE BALL
- parts of the swing, can be pretty tricky to explain, but I'll try my best to explain it, okay so a golf swing is like a pendulum, you swing right to left. Once you lift the club to the right you still look down at the ball. After swimming to the right your have to hit the ball with your right hand strength, in order the ball to fly.
- the way you do your grip defines if the ball will go straight or not. But also the way you swing towards the ball also defines how far the ball is going to go.
Answer:
The ratio of the orbital time periods of A and B is ![\frac{1}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D)
Solution:
As per the question:
The orbit of the two satellites is circular
Also,
Orbital speed of A is 2 times the orbital speed of B
(1)
Now, we know that the orbital speed of a satellite for circular orbits is given by:
![v_{o} = \farc{2\piR}{T}](https://tex.z-dn.net/?f=v_%7Bo%7D%20%3D%20%5Cfarc%7B2%5CpiR%7D%7BT%7D)
where
R = Radius of the orbit
Now,
For satellite A:
![v_{oA} = \farc{2\piR}{T_{a}}](https://tex.z-dn.net/?f=v_%7BoA%7D%20%3D%20%5Cfarc%7B2%5CpiR%7D%7BT_%7Ba%7D%7D)
Using eqn (1):
(2)
For satellite B:
(3)
Now, comparing eqn (2) and eqn (3):
![\frac{T_{a}}{T_{b}} = \farc{1}{2}](https://tex.z-dn.net/?f=%5Cfrac%7BT_%7Ba%7D%7D%7BT_%7Bb%7D%7D%20%3D%20%5Cfarc%7B1%7D%7B2%7D)