Answer:Explained below.
Explanation:
Uranus rings is made up of jet black, coal-like particles in small bands, making them difficult to perceive from Earth.This indicates that they are probably composed of a mixture of the ice and a dark material. The nature of material is dismal, but it might be some organic compounds greatly darkened by the charged particle irradiation from the Uranian magnetosphere. Rings were discovered by using a infrared telescope throughout the occultation of a star as Uranus passed in front of it. The light from the star dimmed many times before it was obstructed by the disk of Uranus and subsequently, showing the presence of various distinct rings.
Answer:
at t=46/22, x=24 699/1210 ≈ 24.56m
Explanation:
The general equation for location is:
x(t) = x₀ + v₀·t + 1/2 a·t²
Where:
x(t) is the location at time t. Let's say this is the height above the base of the cliff.
x₀ is the starting position. At the base of the cliff we'll take x₀=0 and at the top x₀=46.0
v₀ is the initial velocity. For the ball it is 0, for the stone it is 22.0.
a is the standard gravity. In this example it is pointed downwards at -9.8 m/s².
Now that we have this formula, we have to write it two times, once for the ball and once for the stone, and then figure out for which t they are equal, which is the point of collision.
Ball: x(t) = 46.0 + 0 - 1/2*9.8 t²
Stone: x(t) = 0 + 22·t - 1/2*9.8 t²
Since both objects are subject to the same gravity, the 1/2 a·t² term cancels out on both side, and what we're left with is actually quite a simple equation:
46 = 22·t
so t = 46/22 ≈ 2.09
Put this t back into either original (i.e., with the quadratic term) equation and get:
x(46/22) = 46 - 1/2 * 9.806 * (46/22)² ≈ 24.56 m
Answer:
A. Inertial Confinement and B. Magnetic Confinement
Answer:
a)
b)
Explanation:
a)
The width of the central bright in this diffraction pattern is given by:
when m is a natural number.
here:
- m is 1 (to find the central bright fringe)
- D is the distance from the slit to the screen
- a is the slit wide
- λ is the wavelength
So we have:
b)
Now, if we do m=2 we can find the distance to the second minima.
Now we need to subtract these distance, to get the width of the first bright fringe :
I hope it heps you!
