distance to the star Betelgeuse: 640 ly
As we know that
also we know that
So the distance of Betelgeuse = 640 ly
distance to the star VY Canis Majoris:
distance to the galaxy Large Magellanic Cloud: 49976 pc
now we have
distance to Neptune at the farthest: 4.7 billion km
now the order of distance from least to greatest is as following
1. distance to Neptune at the farthest
2. distance of Betelgeuse
3. distance to the star VY Canis Majoris
4. distance to the galaxy Large Magellanic Cloud
The final temperature is 83 K.
<u>Explanation</u>:
For an adiabatic process,
Given:-
(the gas is monoatomic)
T = 275 0.30
T = 83 K.
Answer:
The shell hit at a distance of 1.9 x 10² km
The time of flight of the shell was 5.3 x 10² s
Explanation:
The position of the shell is given by the vector "r":
r = (x0 + v0 * t * cos α ; y0 + v0 * t * sin α + 1/2 g t²)
where:
x0 = initial horizontal position
v0 = magnitude of the initial velocity
t = time
α = launching angle
y0 = initial vertical position
g = acceleration of gravity
When the shell hit, the vertical component (ry) of the vector position r is 0. See figure.
Then:
ry = 0 = y0 + v0 * t * sin α + 1/2 g t²
Since the gun is at the center of our system of reference, y0 and x0 = 0
0 = t (v0 sin α + 1/2 g t)
t= 0 is discarded as solution
v0 sin α + 1/2 g t = 0
t = -2v0 sin α / g
t = (-2 * 2610 m/s * sin 81.9°)/ (-9.8 m/s²) = 5.3 x 10² s. This is the time of flight of the shell until it hit.
Then, the distance at which the shell hit is:
Distance = Module of r = ( x0 + v0 * t * cos α; 0) = x0 + v0 * t * cos α
Distance = 2.61 km/s * 5.3 x 10² s * cos 81.9 = 1.9 x 10² km
