It’s d because it’s useful output which would be light over total input
7.5/8
Explanation:
At any point on the arrow's trajectory, the horizontal component of the velocity is the same. Therefore, the horizontal component of the velocity at the top of its trajectory is
The characteristics of the diffraction phenomenon allow to find the result for the shape of the points of light that you pass the tree is:
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The shape of the dots is circular because it is in the range of far-field diffraction.
Diffraction is the phenomenon where the undulatory part of the light becomes evident, it is the interference of the waves that make up each ray of light, for this phenomenon to occur it must be fulfilled that the wavelength is of the order of the space where pass the light.
In the leafy tree it has many leaves, but there are spaces between them, some of these spaces are small and it fulfills the diffraction condition, therefore we see bright spots and not a continuous shadow.
Diffraction can be classified depending on the distance to the observer:
- Near field or fresnel. In this case the distance from the observer is small and we can see the shape of the object that creates the diffraction.
- Far field or Fraunhoger. In this case the distance between the obstacle (leaves) and the person is great, here the information on the shape of things is lost and we have two observable forms. Lines for the case of slits and circles for the case of objects with a closed shape.
In this case, the distance from the leaves to the observer is large, therefore we are in the case of far-field diffraction and since the edge of the leaves that forms the diffraction is closed, the observable shape is a circle.
In conclusion using the characteristics of the diffraction phenomenon we can find the result for the shape of the points of light that pass the tree is:
-
The shape of the dots is circular because it is in the range of far-field diffraction.
Learn more about diffraction here: brainly.com/question/20140459
Answer:
a) 1.082 × 10⁻¹⁹C ( e = 1.6 × 10⁻¹⁹C)
b) 3.466 × 10¹¹ N/C
Explanation:
a)
p(r) = -A exp ( - 2r/a₀)
Q = ₀∫^∞ ₀∫^π ₀∫^2xπ p(r)dV = -A ₀∫^∞ ₀∫^π ₀∫^2π exp ( - 2r/a₀)r² sinθdrdθd∅
Q = -4πA ₀∫^∞ exp ( - 2r/a₀)r²dr = -e
now using integration by parts;
A = e / πa₀³
p(r) = - (e / πa₀³) exp (-2r/a₀)
Now Net charge inside a sphere of radius a₀ i.e Qnet is;
= e - (e / πa₀³) ₀∫^a₀ ₀∫^π ₀∫^2π r² exp (-2r/a₀)dr
= e - e + 5e exp (-2) = 1.082 × 10⁻¹⁹C ( e = 1.6 × 10⁻¹⁹C)
b)
Using Gauss's law,
E × 4πa₀ ² = Qnet / ∈₀
E = 4πa₀ ² × Qnet × 1/a₀²
E = 3.466 × 10¹¹ N/C
Answer:
When uranium is mined, it consists of approximately 99.3% uranium-238 (U238), 0.7% uranium-235 (U235), and < 0.01% uranium-234 (U234). These are the different uranium isotopes. Isotopes of uranium contain 92 protons in the atom's center or nucleus. (The number of protons in the nucleus is what makes the atoms "uranium.") The U238 atoms contain 146 neutrons, the U235 atoms contain 143 neutrons, and the U234 atoms contain only 142 neutrons. The total number of protons plus neutrons gives the atomic mass of each isotope — that is 238, 235, or 234, respectively. On an atomic level, the size and weight of these isotopes are slightly different. This implies that with the right equipment and under the right conditions, the isotopes can be separated.
Explanation:
