Hi there!
We can use the following (derived) equation to solve for the final velocity given height:
vf = √2gh
We can rearrange to solve for height:
vf² = 2gh
vf²/2g = h
Plug in the given values (g = 9.81 m/s²)
(13)²/2(9.81) = 8.614 m
We can calculate time using the equation:
vf = vi + at, where:
vi = initial velocity (since dropped from rest, = 0 m/s)
a = acceleration (in this instance, due to gravity)
Plug in values:
13 = at
13/a = t
13/9.81 = 1.325 sec
Answer:
there are 25 kg objective travelling at 2m/s to the right.
Answer:
we agree with
Edgar: The net force on the ball at the top position is 9 N. Both the tension and the weight are acting downward so you have to add them.
Explanation:
Weight of the ball is given as
so we have
now tension force at the top is given as
Now at the top position by force equation we can say that ball will have two downwards forces
1) Tension force
2) Weight of the ball
so net force on the ball is given as
So we agree with
Edgar: The net force on the ball at the top position is 9 N. Both the tension and the weight are acting downward so you have to add them.
The ones that interact would be Atmosphere, biosphere, and cryosphere. :)
Good. You can do some very interesting experiments with that equipment.