<u>Answer:</u> The angle of diffraction is 0.498°
<u>Explanation:</u>
To calculate the angle of diffraction, we use the equation given by Bragg, which is:

where,
n = order of diffraction = 3
= wavelength of the light =
(Conversion factor:
)
d = spacing between the crystal planes = 0.100 mm =
(Conversion factor: 1 m = 1000 mm)
= angle of diffraction = ?
Putting values in above equation:

Hence, the angle of diffraction is 0.498°
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.
Answer:
He needs 1.53 seconds to stop the car.
Explanation:
Let the mass of the car is 1500 kg
Speed of the car, v = 20.5 m/s
He will not push the car with a force greater than, 
The impulse delivered to the object is given by the change in momentum as :

So, he needs 1.53 seconds to stop the car. Hence, this is the required solution.
I believe Intermolecular forces hold, <span>molecules, ions, and atoms? But I would see if that doesn't sound familiar check it with a site or something?</span>
While plane is moving under tailwind condition it took time "t"
so here we will have

here net speed of the plane will be given as


similarly when it moves under the condition of headwind its net speed is given as

now time taken to cover the distance is 2 hours more

now solving two equations

solving above for v_w we got
