Answer :
Velocity will be ![3.28\times 10^{-11}m/sec](https://tex.z-dn.net/?f=3.28%5Ctimes%2010%5E%7B-11%7Dm%2Fsec)
Explanation:
We have given glass surface has a diameter of 1.5 mm
And charge q = 1.60 nC
Radius of electrons orbit r = height of electron above surface + radius of sphere = ![=1.6+\frac{1.5}{2}=2.35mm = 0.00235m](https://tex.z-dn.net/?f=%3D1.6%2B%5Cfrac%7B1.5%7D%7B2%7D%3D2.35mm%20%3D%200.00235m)
Force on electron is given by
, here q is charge on sphere and e is charge on electron
![F=\frac{1}{4\pi \epsilon _0}\frac{qe}{r^2}=\frac{kqe}{r^2}=\frac{9\times 10^9\times 1.6\times 10^{-9}\times 1.6\times 10^{-19}}{0.00235^2}=4.172\times 10^{-13}N](https://tex.z-dn.net/?f=F%3D%5Cfrac%7B1%7D%7B4%5Cpi%20%5Cepsilon%20_0%7D%5Cfrac%7Bqe%7D%7Br%5E2%7D%3D%5Cfrac%7Bkqe%7D%7Br%5E2%7D%3D%5Cfrac%7B9%5Ctimes%2010%5E9%5Ctimes%201.6%5Ctimes%2010%5E%7B-9%7D%5Ctimes%201.6%5Ctimes%2010%5E%7B-19%7D%7D%7B0.00235%5E2%7D%3D4.172%5Ctimes%2010%5E%7B-13%7DN)
This force work as centripetal force
So ![F=\frac{mv^2}{r}](https://tex.z-dn.net/?f=F%3D%5Cfrac%7Bmv%5E2%7D%7Br%7D)
![4.172\times 10^{-13}=\frac{9.11\times 10^{-31}v^2}{0.00235}](https://tex.z-dn.net/?f=4.172%5Ctimes%2010%5E%7B-13%7D%3D%5Cfrac%7B9.11%5Ctimes%2010%5E%7B-31%7Dv%5E2%7D%7B0.00235%7D)
v = ![=0.0328\times 10^{-9}=3.28\times 10^{-11}m/sec](https://tex.z-dn.net/?f=%3D0.0328%5Ctimes%2010%5E%7B-9%7D%3D3.28%5Ctimes%2010%5E%7B-11%7Dm%2Fsec)
Saturn's majestic rings
This majestic image of Saturn was taken by the Cassini spacecraft as it passed through the shadow of the giant planet. The rings are so reflective, they appear to light up the night side of the planet. Just above the main ring system on the left is a tiny, unsuspecting blue dot, ultimately responsible for this image: Earth.
(click to enlarge)
Every amateur astronomer knows how spectacular Saturn's rings look from even the smallest telescope. The glowing bands and Cassini Division can even be resolved through a set of handheld binoculars. For hundreds of years after Galileo first observed them in 1610, scientists believed Saturn to be the only planet with rings.
That myth wasn't dispelled until 1977 when a star passed behind the planet Uranus in an event called stellar occultation. To scientists' surprise, the star blinked on and off 9 times before passing behind the surface of the planet. Even though the rings weren't visible with the naked eye the stellar occultation indicated that something was there to block the light. Even more surprising to scientists was the discovery made by Voyager 1 in 1979 that Jupiter had rings. Now it's up to Neptune to complete the story.
Uranus' Rings
The rings of Uranus seen here have been digitally brightened. The particles are primarily chunks of ice that have been darkened by rocks, and actually reflect as little light as charcoal.
(click to enlarge)
In the mid-1980s, there was a stellar occultation of Neptune and, just like Uranus, the starlight blinked before reaching the planet. Scientists weren't convinced that Neptune's rings completely circled the planet, but instead were smaller arcs. It wasn't until Voyager 2 passed by in 1989 that the truth became known. Neptune had a complete ring system as well.
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Answer:
The X-axis component of the vector from the point
to
would be
. While the Y-axis component would be
.
A vector in a two-dimensional space from point
to
can be obtained as the vectorial subtraction between the endpoint and the initial point:
![\vec{V}=(x_c-x_d,y_c-y_d)](https://tex.z-dn.net/?f=%5Cvec%7BV%7D%3D%28x_c-x_d%2Cy_c-y_d%29)
Explanation:
It is given that,
Force on grindstone, F = 180 N
Radius of grindstone, r = 0.28 m
Mass of grindstone, m = 75 kg
We need to find the angular acceleration of the grindstone. In rotational motion, the relation between the torque and angular acceleration is given by :
![\tau=I\times \alpha](https://tex.z-dn.net/?f=%5Ctau%3DI%5Ctimes%20%5Calpha)
![\alpha=\dfrac{\tau}{I}](https://tex.z-dn.net/?f=%5Calpha%3D%5Cdfrac%7B%5Ctau%7D%7BI%7D)
I is the moment of inertia of solid disk, ![I=\dfrac{1}{2}mr^2](https://tex.z-dn.net/?f=I%3D%5Cdfrac%7B1%7D%7B2%7Dmr%5E2)
is the torque exerted, ![\tau=F\times r](https://tex.z-dn.net/?f=%5Ctau%3DF%5Ctimes%20r)
![\alpha=\dfrac{F\times r}{\dfrac{1}{2}mr^2}](https://tex.z-dn.net/?f=%5Calpha%3D%5Cdfrac%7BF%5Ctimes%20r%7D%7B%5Cdfrac%7B1%7D%7B2%7Dmr%5E2%7D)
![\alpha=\dfrac{2F}{mr}](https://tex.z-dn.net/?f=%5Calpha%3D%5Cdfrac%7B2F%7D%7Bmr%7D)
![\alpha=\dfrac{2\times 180\ N}{75\ kg\times 0.28\ m}](https://tex.z-dn.net/?f=%5Calpha%3D%5Cdfrac%7B2%5Ctimes%20180%5C%20N%7D%7B75%5C%20kg%5Ctimes%200.28%5C%20m%7D)
![\alpha =17.14\ rad/s^2](https://tex.z-dn.net/?f=%5Calpha%20%3D17.14%5C%20rad%2Fs%5E2)
So, the angular acceleration of the disk is
. Hence, this is the required solution.