The answer is A.
When a substance is a gas, it fills its container no matter how many particles it's made of, however the number of particles present will directly affect the pressure and temperature of the system.
Answer:
The grating spacing of the beetle is 
Explanation:
The concept to solve this problem is relate to interference effect given in the Young's Slits. Here was demonstrated that the length of the side labelled \lambda is known as the path difference. The equation is given by,
Where,
= wavelenght of light
N = a positive integer: 1,2,3...
= Angle from the center of the wall to the dark spot
d= width of the slit
Replacing our values we have that for n=1,
Therefore the grating spacing of the beetle is
Answer:
a) The student feel light
b) Nbottom = 758 N
c) N'top= 236 N
d) N'bottom= 1055 N
Explanation:
a) W= 659N , Ntop= 560N
W > Ntop ---> Student feel less weight
b) Top:
∑F= W - Ntop = m.v²/R
m.v²/R = 659N - 560 N = 99 N
Bottom:
∑F= Nbottom- W = m.v²/R
Nbottom= W + m.v²/R = 659N + 99 N = 758N
c) W= 659 N , Ntop= 560 N , v'=2.v
N'top= ?
∑F= W - N'top = m.v'²/R
N'top= W - 4.m.v²/R
N'top = 659 N - 4. 99 N = 263 N
d) N'bottom = ?
∑Fbottom= N'bottom- W = m.v'²/R
N'bottom = W + 4.m.v²/R = 659 N + 4. 99 N = 1055 N
It's just asking you to sit down and COUNT the little squares in each sector.
It'll help you keep everything straight if you take a very sharp pencil and make a tiny dot in each square as you count it. That way, you'll be able to see which ones you haven't counted yet, and also you won't count a square twice when you see that it already has a dot in it.
(If, by some chance, this is a picture of the orbit of a planet revolving around the sun ... as I think it might be ... then you should find that both sectors jhave the same number of squares.)
Surface miners work aboveground. (Apex) ^-^
