Answer: The height (position) of the ball and the acceleration due gravity
Explanation:
In this case we are taking about gravitational potential energy, which is the energy a body or object possesses, due to its position in a gravitational field. In this sense, this energy depends on the relative height of an object with respect to some point of reference and associated with the gravitational force.
In the case of the Earth, in which the gravitational field is considered constant, the gravitational potential energy 
will be:
Where:
is the mass of the ball
is the acceleration due gravity (assuming the ball is on the Earth surface)
is the height (position) of the ball respect to a given point
Note the value of the gravitational potential energy is directly proportional to the height.
An activity that is relatively short in time <10 seconds and has few repetitions predominantly uses the ATP/PC energy system. The cellular respiration procedure that changes food energy into ATP which is a form of energy is largely reliant on oxygen obtainability. During exercise the source and request of oxygen obtainable to muscle is unnatural by period and strength and by the individual’s cardiorespiratory suitability level.
Steps of the ATP-PC system:
1. Primarily, ATP kept in the myosin cross-bridges which is microscopic contractile parts of muscle is broken down to issue energy for muscle shrinkage. This action consents the by-products of ATP breakdown which are the adenosine diphosphate and one single phosphate all on its own.
2. Phosphocreatine is then broken down by the enzyme creatine kinase into creatine and phosphate.
3. The energy free in the breakdown of PC permits ADP and Pi to rejoin creating more ATP. This newly made ATP can now be broken down to issue energy to fuel activity.
Answer: A
Explanation:
Well the high and lows effect the humidity the more humidity the more hot it is so the high brings higher temperatures.
Answer:
<em>The athlete will rise 1.10 meters off the ground</em>
Explanation:
<u>Vertical Motion</u>
If an object is launched vertically upwards at an initial speed vo, then it will reach a maximum height given by
The athlete can exert a net force upwards equal to twice his weight. It makes him accelerate upwards at
The speed at the end of his push can be computed by
Replacing the value of a obtained above:
where y is the length of this crouch
This is the initial speed of this vertical launch, thus
Answer:
The acceleration of the cart is 1.0 m\s^2 in the negative direction.
Explanation:
Using the equation of motion:
Vf^2 = Vi^2 + 2*a*x
2*a*x = Vf^2 - Vi^2
a = (Vf^2 - Vi^2)/ 2*x
Where Vf is the final velocity of the cart, Vi is the initial velocity of the cart, a the acceleration of the cart and x the displacement of the cart.
Let x = Xf -Xi
Where Xf is the final position of the cart and Xi the initial position of the cart.
x = 12.5 - 0
x = 12.5
The cart comes to a stop before changing direction
Vf = 0 m/s
a = (0^2 - 5^2)/ 2*12.5
a = - 1 m/s^2
The cart is decelerating
Therefore the acceleration of the cart is 1.0 m\s^2 in the negative direction.
