66666666666666666666666666666666666666666666666666666666666666666666666666666666666666
Answer:
Associations Clusters
Explanation:
There is three basic type of clusters, globular clusters, open clusters, associations clusters. Associations cluster consist of younger stars that is younger than globular and open clusters stars.
Great Question! I happened to be a physics nerd!
Answer:
C. Two hydrogen nuclei, each with only one proton, fuse to form deuterium, a form of hydrogen with one proton.
MAKE SURE TO SEE EXPLANATION!
Explanation:
In the core of the Sun, or any other main sequence star, there is no single fusion process. Instead, complex sequences of processes occur to make helium nuclei from hydrogen nuclei (i.e. protons). The proton-proton chain provides for the majority of energy generation in stars with masses less than that of the Sun. One difficulty in creating a helium nucleus (two protons and two neutrons) is that there are only protons to begin with. Some protons must be turned into neutrons in some way. The first step is to combine two protons to form a deuterium nucleus (also known as a deuteron). That's a hefty hydrogen nucleus with one proton and one neutron. Such a proton-proton contact is highly unlikely, and it has never been detected in a laboratory. Fortunately, the Sun's core is incredibly hot and dense, with an incredible number of protons packed inside. Even a low likelihood event will occur every now and again. Along with each deuteron, a positron (an "anti-electron") and a neutrino are created. Because the Sun's core is plasma, there are a lot of free electrons, thus the positron doesn't live long until it and an electron collide and annihilate, resulting in gamma radiation. The deuteron then interacts with a proton to form a helium 3 nucleus. That is a high-probability interaction, and it occurs swiftly. Two helium 3 nuclei join in the third phase to generate a helium 4 ("regular" helium) nucleus and a proton. Branch I of the proton-proton (p-p) chain is responsible for this. Another stage is required because reactions between helium 3 and helium 4 nuclei are possible. There are two conceivable reactions (named Branch II and Branch III), and I'll save you the gory details. It gets much more complicated since theoretical calculations indicate that a reaction between a helium 3 nucleus and a proton is feasible — Branch IV. This reaction has an incredibly low likelihood of occurring, far lower than the Branch I reaction, thus it must be exceedingly rare. The Carbon-Nitrogen-Oxygen (CNO) Cycle is another method for reducing hydrogen to helium. It does not generate much energy in the Sun, but it is the principal energy generation mechanism in larger stars.
The maximum velocity in a banked road, ignoring friction, is given by;
v = Sqrt (Rg tan ∅), where R = Radius of the curved road = 2*1000/2 = 1000 m, g = gravitational acceleration = 9.81 m/s^2, ∅ = Angle of bank.
Substituting;
30 m/s = Sqrt (1000*9.81*tan∅)
30^2 = 1000*9.81*tan∅
tan ∅ = (30^2)/(1000*9.81) = 0.0917
∅ = tan^-1(0.0917) = 5.24°
Therefore, the road has been banked at 5.24°.
Compression is above the equilibrium and rarefaction is below
