Answer:
hmax = 1/2 · v²/g
Explanation:
Hi there!
Due to the conservation of energy and since there is no dissipative force (like friction) all the kinetic energy (KE) of the ball has to be converted into gravitational potential energy (PE) when the ball comes to stop.
KE = PE
Where KE is the initial kinetic energy and PE is the final potential energy.
The kinetic energy of the ball is calculated as follows:
KE = 1/2 · m · v²
Where:
m = mass of the ball
v = velocity.
The potential energy is calculated as follows:
PE = m · g · h
Where:
m = mass of the ball.
g = acceleration due to gravity (known value: 9.81 m/s²).
h = height.
At the maximum height, the potential energy is equal to the initial kinetic energy because the energy is conserved, i.e, all the kinetic energy was converted into potential energy (there was no energy dissipation as heat because there was no friction). Then:
PE = KE
m · g · hmax = 1/2 · m · v²
Solving for hmax:
hmax = 1/2 · v² / g
Answer:
Your answer will be 600meters
Well you’d have a force due to gravity, the normal force which will be perpendicular to the sources (meaning you’ll have components to this vector), and you’d have the force of friction opposing the motion of the box. I’m also assuming there’s no air resistance. In this case you’d have three vector forces.
The density of an object can be calculated using the formula Density = Mass/Volume. In this case however we are searching for the volume and must rearrange the formula so that we are solving for the volume. If you multiply both sides by volume and then divide both sides by mass you end up with the equation Volume = Mass/Density.
Volume = 1500g/1.5g/cm^3
Volume = 1000 cm^3
What happens to end a of the rod when the ball approaches it closely this first time is; It is strongly attracted.
<h3>Electrostatics</h3>
I have attached the image of the rod.
We are told that the ball is much closer to the end of the rod than the length of the rod. Thus, if we point down the rod several times, the distance of approach will experience no electric field and as such the charge on end point A of the rod must be comparable in magnitude to the charge on the ball.
This means that their fields will cancel.
Finally, we can conclude that when a charge is brought close to a conductor, the opposite charges will all navigate to the point that is closest to the charge and as a result, a strong attraction will be created.
This also applies to a strong conducting rod and therefore it is strongly attracted.
Read more about Electrostatics at; brainly.com/question/18108470