No. The moon always keeps the same side facing us. Its rotation and revolution periods are equal.
Wow ! This will take more than one step, and we'll need to be careful
not to trip over our shoe laces while we're stepping through the problem.
The centripetal acceleration of any object moving in a circle is
(speed-squared) / (radius of the circle) .
Notice that we won't need to use the mass of the train.
We know the radius of the track. We don't know the trains speed yet,
but we do have enough information to figure it out. That's what we
need to do first.
Speed = (distance traveled) / (time to travel the distance).
Distance = 10 laps of the track. Well how far is that ? ? ?
1 lap = circumference of the track = (2π) x (radius) = 2.4π meters
10 laps = 24π meters.
Time = 1 minute 20 seconds = 80 seconds
The trains speed is (distance) / (time)
= (24π meters) / (80 seconds)
= 0.3 π meters/second .
NOW ... finally, we're ready to find the centripetal acceleration.
<span> (speed)² / (radius)
= (0.3π m/s)² / (1.2 meters)
= (0.09π m²/s²) / (1.2 meters)
= (0.09π / 1.2) m/s²
= 0.236 m/s² . (rounded)
If there's another part of the problem that wants you to find
the centripetal FORCE ...
Well, Force = (mass) · (acceleration) .
We know the mass, and we ( I ) just figured out the acceleration,
so you'll have no trouble calculating the centripetal force. </span>
To answer, evaluate the power of 10 in the given choices. If it is positve, move the decimal n places to the right. If it is negative, move the decimal n corresponding places to the left. From all the choices given, only the choices D, E, and F will give us the correct answer.
Galileo Galilei was the first scientist to perform experiments in order to test his ideas. He was also the first astronomer to systematically observe the skies with a telescope.
:)
Explanation:
It is given that,
An electron is released from rest in a weak electric field of, ![E=2.3\times 10^{-10}\ N/C](https://tex.z-dn.net/?f=E%3D2.3%5Ctimes%2010%5E%7B-10%7D%5C%20N%2FC)
Vertical distance covered, ![s=1\ \mu m=10^{-6}\ m](https://tex.z-dn.net/?f=s%3D1%5C%20%5Cmu%20m%3D10%5E%7B-6%7D%5C%20m)
We need to find the speed of the electron. Let its speed is v. Using third equation of motion as :
![v^2-u^2=2as](https://tex.z-dn.net/?f=v%5E2-u%5E2%3D2as)
.............(1)
Electric force is
and force of gravity is
. As both forces are acting in downward direction. So, total force is:
![F=mg+qE](https://tex.z-dn.net/?f=F%3Dmg%2BqE)
![F=9.1\times 10^{-31}\times 9.8+1.6\times 10^{-19}\times 2.3\times 10^{-10}](https://tex.z-dn.net/?f=F%3D9.1%5Ctimes%2010%5E%7B-31%7D%5Ctimes%209.8%2B1.6%5Ctimes%2010%5E%7B-19%7D%5Ctimes%202.3%5Ctimes%2010%5E%7B-10%7D)
![F=4.57\times 10^{-29}\ N](https://tex.z-dn.net/?f=F%3D4.57%5Ctimes%2010%5E%7B-29%7D%5C%20N)
Acceleration of the electron, ![a=\dfrac{F}{m}](https://tex.z-dn.net/?f=a%3D%5Cdfrac%7BF%7D%7Bm%7D)
![a=\dfrac{4.57\times 10^{-29}\ N}{9.1\times 10^{-31}\ kg}](https://tex.z-dn.net/?f=a%3D%5Cdfrac%7B4.57%5Ctimes%2010%5E%7B-29%7D%5C%20N%7D%7B9.1%5Ctimes%2010%5E%7B-31%7D%5C%20kg%7D)
![a=50.21\ m/s^2](https://tex.z-dn.net/?f=a%3D50.21%5C%20m%2Fs%5E2)
Put the value of a in equation (1) as :
![v=\sqrt{2as}](https://tex.z-dn.net/?f=v%3D%5Csqrt%7B2as%7D)
![v=\sqrt{2\times 50.21\times 10^{-6}}](https://tex.z-dn.net/?f=v%3D%5Csqrt%7B2%5Ctimes%2050.21%5Ctimes%2010%5E%7B-6%7D%7D)
v = 0.010 m/s
So, the speed of the electron is 0.010 m/s. Hence, this is the required solution.