Answer:
0.57 m
Explanation:
First of all, we need to calculate the time it takes for the ball to cover the horizontal distance between the starting position and the crossbar. This can be done by analzying the horizontal motion only. In fact, the horizontal velocity is constant and it is
And the distance to cover is
d = 19 m
So the time taken is
Now we want to find how high the ball is at that time. The initial vertical velocity is
So the vertical position of the ball at time t is
where g = 9.8 m/s^2 is the acceleration of gravity. Substituting t = 2.04 s, we find
The crossbar height is 3.05 m, so the difference is
So the ball passes 0.57 m above the crossbar.
6.0 is the answer. hope this helps ya
the correct answer is unbalanced because if it was balanced it would be floating because it would have no push or pull force.
The question is really specific enough, but I will take a guess. If nothing else changes (like the mass or the gravitational constant) then the Potential Energy will increase with an increase in height. Ep will decrease with a decrease in height.
If you are talking about Ke, then the height does not matter unless the object is moving up. Then it slows down. When that happens the Ke decreases because the formula for Ke = 1/2 m v^2. If v becomes less then Ke becomes less.
I suppose you could say if the object is going down and the distance is decreasing then the speed will increase. It's an inverse relationship.
I think you are best talking about Pe.
Answer:
35.6 N
Explanation:
We can consider only the forces acting along the horizontal direction to solve the problem.
There are two forces acting along the horizontal direction:
- The horizontal component of the pushing force, which is given by
with 
- The frictional force, whose magnitude is
where 
, m=8.2 kg and g=9.8 m/s^2.
The two forces have opposite directions (because the frictional force is always opposite to the motion), and their resultant must be zero, because the suitcase is moving with constant velocity (which means acceleration equals zero, so according to Newton's second law: F=ma, the net force is zero). So we can write:
