Explanation:
a) The height of the ball h with respect to the reference line is
so its initial gravitational potential energy 
is
b) To find the speed of the ball at the reference point, let's use the conservation law of energy:
We know that the initial kinetic energy 
as well as its final gravitational potential energy 
are zero so we can write the conservation law as
Note that the mass gets cancelled out and then we solve for the velocity v as
A star is born when clouds of dust and elements are gathered together in a certain space due to gravity, more and more mass and therefore pressure builds. When the pressure becomes enough to overcome the electronic repulsive force between two hydrogen nuclei, they are forced together and massive amounts of energy are given off forming helium atoms. This energy is then used to fuse other nuclei together. This could be compared to the way human life starts, where instead of 2 nuclei joining together to start a life cycle, two gametes, or sex cells are joined together. Also at the start of both a star and persons life, we are weak and we gain strength until we reach the height of our existence, then humans slowly become less efficient at doing what they do until eventually they cannot sustain themselves any further.
Answer:
0
Explanation:
m = Mass of person
g = Acceleration due to gravity = 9.81 m/s²
d = Vertical height from the ground
F = Force = Weight = mg
Net work done would be
Hence, the work done on the person by the gravitational force is 0
Answer:
Increases, increases
Explanation:
The current is directly proportional to the voltage and inversely proportional to the resistance. The implication of this is that, whenever the voltage is increased, the current increases simultaneously. On the other hand, if the resistance is increased, the current will decrease accordingly and vice versa.
Recall that power is given by P= V^2/R where;
P= power, V= voltage and R= resistance
We can see that power and resistance are inversely related hence decreasing the resistance increases the power output of the lightbulb.
