Answer:
7.89 7.91
Explanation:
The ranges of measurement lie between 7.92-0.05 and 7.92+0.05
7.87g and 7.97g
Answer: sheet of charge
Explanation:
a )
Since the charge is negative , potential will be negative near it . At a far point potential will be less negative. So potential will virtually increase on going away from the sheet . At infinity it will become almost zero. Electric field will be towards the plate , so potential will decrease towards the plate.
b ) The shape of equi -potential surface will be plane parallel to the sheet of charge because electric field will be perpendicular to the sheet of charge and almost uniform near the sheet of charge. The equi- potential surface is always perpendicular to electric field.
C ) Electric field which is almost uniform near the sheet of charge is equal t the following
E = σ / ε₀ where σ is charge density of surface and ε₀ is permittivity of medium whose value is 8.85 x 10⁻¹²
E = 3 x 10⁻⁹ / 8.85 x 10⁻¹²
= .3389 x 10³
= 338.9 V / m
spacing between 1 V
= 1 / 338.9 m
= 2.95 X 10⁻3 m
= 2.95 mm.
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.
A is the answer you can believe me
increased with an increased current flow
