Given parameters:
Mass of the car = 1000kg
Unknown:
Height = ?
To find the heights for the different amount potential energy given, we need to understand what potential energy is.
Potential energy is the energy at rest due to the position of a body.
It is mathematically expressed as:
P.E = mgh
m is the mass
g is the acceleration due to gravity = 9.8m/s²
h is the height of the car
Now the unknown is h, height and we make it the subject of the expression to make for easy calculation.
h = 
<u>For 2.0 x 10³ J;</u>
h =
= 0.204m
<u>For 2.0 x 10⁵ J;</u>
h =
= 20.4m
<u>For 1.0kJ = 1 x 10³J; </u>
h =
= 0.102m
The angle that the cart rolls with the horizontal. The closer the ramp gets to 90 degrees the faster the cart will accelerate.
<span>the statement that is true regarding flexibility is : b. a joint's range of motion will be lost if the joint is not used regularly.
Our body is like a machine. If we not constantly heat it up, our body will be more prone to injury. We can see that the old people who lived within the tribe in the middle of the mountain are far stronger than the one who lived in the city.</span>
X- points away from the charge
y- points in the direction of the force on the positive charge
z- points toward the charge
To solve this problem we will apply the linear motion kinematic equations. With the data provided we will calculate the time of the first object to fall. Later we will get the time difference between the two. This difference will allow us to find the free fall distance. Through the distance we will find the initial velocity, that is,



The second object is thrown downward at one second later and it meets the first object at the water is


The distance of the object will travel due to free fall acceleration is



The distance of the object will travel due to its initial velocity is




Therefore the initial speed of the second object is 21.06m/s