Here,
height at failure, h1 = 525 m,
upward acceleration, a = 2.25 m/s^2,
velocity = v m/s,
<span>
SO, </span>
<span>
v^2 = 2*a*h = 2*2.25*525 = 2362.5 </span>
Now, acceleration, g = 9.8 m/s^2,
<span>
SO, </span>
<span>
heigt, h1 = v^2/2g = 2362.5 / 2*9.8 = 120.54 meters </span>
Hence,
<span>
a) </span>
Total height = 525+120.54 = 645.54 meters
b)
<span>time, for h1, t = v/g = sqrt(2362.5)/9.8 = 4.96 sec
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Modelling the structure of the atom is important because modeling replaces the real system with something similar but easier to examine. Option B
<h3>What is modeling?</h3>
A model is a representation of reality. We know that a model could help us to recreate reality in a manner that we could be able to relate fully with it. A model could be used also a means of explanation.
The atomic models that we have usually help us to understand more abut the atom. Therefore, modelling the structure of the atom is important because modeling replaces the real system with something similar but easier to examine. Option B
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Thank you for your question, what you say is true, the gravitational force exerted by the Earth on the Moon has to be equal to the centripetal force.
An interesting application of this principle is that it allows you to determine a relation between the period of an orbit and its size. Let us assume for simplicity the Moon's orbit as circular (it is not, but this is a good approximation for our purposes).
The gravitational acceleration that the Moon experience due to the gravitational attraction from the Earth is given by:
ag=G(MEarth+MMoon)/r2
Where G is the gravitational constant, M stands for mass, and r is the radius of the orbit. The centripetal acceleration is given by:
acentr=(4 pi2 r)/T2
Where T is the period. Since the two accelerations have to be equal, we obtain:
(4 pi2 r) /T2=G(MEarth+MMoon)/r2
Which implies:
r3/T2=G(MEarth+MMoon)/4 pi2=const.
This is the so-called third Kepler law, that states that the cube of the radius of the orbit is proportional to the square of the period.
This has interesting applications. In the Solar System, for example, if you know the period and the radius of one planet orbit, by knowing another planet's period you can determine its orbit radius. I hope that this answers your question.
Answer:
y = 10.2 m
Explanation:
It is given that,
Charge, 
It is placed at a distance of 9 cm at x axis
Charge, 
It is placed at a distance of 16 cm at x axis
We need to find the point on the y-axis where the electric potential zero. The net potential on y-axis is equal to 0. So,

Here,

So,

Squaring both sides,

So, at a distance of 10.2 m on the y axis the electric potential equals 0.