Q1. The answer is 8.788 m/s
V2 = V1 + at
V1 - the initial velocity
V2 - the final velocity
a - the acceleration
t - the time
We have:
V1 = 4.7 m/s
a = 0.73 m/s²
t = 5.6 s
V2 = ?
V2 = 4.7 + 0.73 * 5.6
V2 = 4.7 + 4.088
V2 = 8.788 m/s
Q2. The answer is 9.22 s
V2 = V1 + at
V1 - the initial velocity
V2 - the final velocity
a - the acceleration
t - the time
We have:
V2 = 0 (because it reaches a complete stop)
V1 = 4.7 m/s
a = -0.51 m/s²
t = ?
0 = 4.7 + (-0.51)*t
0 = 4.7 - 0.51t
0.51t = 4.7
t = 4.7 / 0.51
t = 9.22 s
Answer:
3 a is the ans i think so ....
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.
<h2><u>Hello </u><u>There</u></h2>
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It is science, art and occupation of cultivating the soil, producing crops and livestock. It is systematic and controlled use of living organisms and the environment to improve the human condition.
<h3>Hope This Helps</h3>
