Answer:
200N
Explanation:
mass(m) = 10 kg
acceleration(a) = 20 m/s^2
Force = mass * acceleration
= 10*20
= 200 N
Force = 200N
Answer:
225 rpm
Explanation:
The angular acceleration of the fan is given by:
where
is the final angular speed
is the initial angular speed
is the time interval
For the fan in this problem,
Substituting,
Now we can find the angular speed of the fan at the end of the 5th second, so after t = 5 s. It is given by:
where
Substituting,
Answer:
Given that
speed u=4*10^6 m/s
electric field E=4*10^3 N/c
distance b/w the plates d=2 cm
basing on the concept of the electrostatices
now we find the acceleration b/w the plates to find the horizontal distance traveled by the electron when it hits the plate.
acceleration a=qE/m=
=
m/s
now we find the horizontal distance traveled by electrons hit the plates
horizontal distance
![X=u[2y/a]^{1/2}](https://tex.z-dn.net/?f=X%3Du%5B2y%2Fa%5D%5E%7B1%2F2%7D)
=![4*10^6[2*2*10^{-2}/7*10^{14}]^{1/2}](https://tex.z-dn.net/?f=4%2A10%5E6%5B2%2A2%2A10%5E%7B-2%7D%2F7%2A10%5E%7B14%7D%5D%5E%7B1%2F2%7D)
=
= 3 cm
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.
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