Answer:
0.6983 m/s
Explanation:
k = spring constant of the spring = 0.4 N/m
L₀ = Initial length = 11 cm = 0.11 m
L = Final length = 27 cm = 0.27 m
x = stretch in the spring = L - L₀ = 0.27 - 0.11 = 0.16 m
m = mass of the mass attached = 0.021 kg
v = speed of the mass
Using conservation of energy
Kinetic energy of mass = Spring potential energy
(0.5) m v² = (0.5) k x²
m v² = k x²
(0.021) v² = (0.4) (0.16)²
v = 0.6983 m/s
Before Pluto was discovered, it was predicted. Astronomers had observed that massive objects can affect the orbits of its neighbors, and, after seeing deviations in the orbits of Uranus and Neptune, assumed something substantial existed beyond their orbits.
When Pluto was spotted, it was thought to be the predicted object and was identified as a ninth planet.
A few decades later, astronomers started discovering more and more objects around other stars and didn’t know whether to call them planets or not. There appeared to be a need to define what a planet means, and that led to what some people consider Pluto’s demotion to a dwarf planet.
The International Astronomical Union decided that full-sized planets must orbit the sun, have a round shape, and have cleared their orbits of other objects. Pluto fulfills the first two criteria, but not the third.
It still goes around the sun, it’s round enough, it’s got moons, and behaves like a planet, but the idea is that Pluto did not form the same way as the rest of the planets. Pluto’s orbit is both eccentric and inclined more than the rest of the planets by about 17 degrees. That’s suggests something is different about this object.
This debate about whether to call it a planet or not is silly, because it doesn’t matter to Pluto what you call it. It is an interesting object, goes around the sun, and shows geology and an atmosphere.
There’s a tendency to define objects based on what they are now, but nothing is constant in the universe. There are some issues with the nomenclature, and a definition today may not apply to the same object tomorrow.
Answer:
a
The number of fringe is z = 3 fringes
b
The ratio is 
Explanation:
a
From the question we are told that
The wavelength is 
The distance between the slit is 
The width of the slit is 
let z be the number of fringes that appear between the first diffraction-envelope minima to either side of the central maximum in a double-slit pattern is and this mathematically represented as
Substituting values
z = 3 fringes
b
From the question we are told that the order of the bright fringe is n = 3
Generally the intensity of a pattern is mathematically represented as
![I = I_o cos^2 [\frac{\pi d sin \theta}{\lambda} ][\frac{sin (\pi a sin \frac{\theta}{\lambda } )}{\pi a sin \frac{\theta}{\lambda} } ]](https://tex.z-dn.net/?f=I%20%3D%20I_o%20cos%5E2%20%5B%5Cfrac%7B%5Cpi%20d%20sin%20%5Ctheta%7D%7B%5Clambda%7D%20%5D%5B%5Cfrac%7Bsin%20%28%5Cpi%20a%20sin%20%5Cfrac%7B%5Ctheta%7D%7B%5Clambda%20%7D%20%29%7D%7B%5Cpi%20a%20sin%20%5Cfrac%7B%5Ctheta%7D%7B%5Clambda%7D%20%7D%20%5D)
Where 
is the intensity of the central fringe
And Generally 
![I = I_o co^2 [ \frac{\pi (\frac{n \lambda}{d} )}{\lambda} ] [\frac{\frac{sin (\pi a (\frac{n \lambda}{d} ))}{\lambda} }{\frac{\pi a (\frac{n \lambda}{d} )}{\lambda} } ]](https://tex.z-dn.net/?f=I%20%3D%20I_o%20co%5E2%20%5B%20%5Cfrac%7B%5Cpi%20%28%5Cfrac%7Bn%20%5Clambda%7D%7Bd%7D%20%29%7D%7B%5Clambda%7D%20%5D%20%5B%5Cfrac%7B%5Cfrac%7Bsin%20%28%5Cpi%20a%20%28%5Cfrac%7Bn%20%5Clambda%7D%7Bd%7D%20%29%29%7D%7B%5Clambda%7D%20%7D%7B%5Cfrac%7B%5Cpi%20a%20%28%5Cfrac%7Bn%20%5Clambda%7D%7Bd%7D%20%29%7D%7B%5Clambda%7D%20%7D%20%5D)
![I = I_o cos^2 (n \pi)[\frac{\frac{sin(\pi a (\frac{n \lambda}{d} ))}{\lambda} )}{ \frac{ \pi a (\frac{n \lambda }{d} )}{\lambda} } ]](https://tex.z-dn.net/?f=I%20%3D%20I_o%20cos%5E2%20%28n%20%5Cpi%29%5B%5Cfrac%7B%5Cfrac%7Bsin%28%5Cpi%20a%20%28%5Cfrac%7Bn%20%5Clambda%7D%7Bd%7D%20%29%29%7D%7B%5Clambda%7D%20%29%7D%7B%20%5Cfrac%7B%20%5Cpi%20a%20%28%5Cfrac%7Bn%20%5Clambda%20%7D%7Bd%7D%20%29%7D%7B%5Clambda%7D%20%7D%20%5D)
![I = I_o cos^2 (3 \pi) [\frac{sin (\frac{3 \pi }{6} )}{\frac{3 \pi}{6} } ]](https://tex.z-dn.net/?f=I%20%3D%20I_o%20cos%5E2%20%283%20%5Cpi%29%20%5B%5Cfrac%7Bsin%20%28%5Cfrac%7B3%20%5Cpi%20%7D%7B6%7D%20%29%7D%7B%5Cfrac%7B3%20%5Cpi%7D%7B6%7D%20%7D%20%5D)
Well if it was traveling for an hour then the answer is 8 miles.
