Answer:
f = 19,877 cm and P = 5D
Explanation:
This is a lens focal length exercise, which must be solved with the optical constructor equation
1 / f = 1 / p + 1 / q
where f is the focal length, p is the distance to the object and q is the distance to the image.
In this case the object is placed p = 25 cm from the eye, to be able to see it clearly the image must be at q = 97 cm from the eye
let's calculate
1 / f = 1/97 + 1/25
1 / f = 0.05
f = 19,877 cm
the power of a lens is defined by the inverse of the focal length in meters
P = 1 / f
P = 1 / 19,877 10-2
P = 5D
10^9 giga, 10^6 mega, 10^3 kilo, 10^-3 milli, 10^-6 micro, 10^-9 nano, 10^-12 pico
Potentially they might want centi which is 10^-2
Explanation:
Suppose you want to shine a flashlight beam down a long, straight hallway. Just point the beam straight down the hallway -- light travels in straight lines, so it is no problem. What if the hallway has a bend in it? You could place a mirror at the bend to reflect the light beam around the corner. What if the hallway is very winding with multiple bends? You might line the walls with mirrors and angle the beam so that it bounces from side-to-side all along the hallway. This is exactly what happens in an optical fiber.
The light in a fiber-optic cable travels through the core (hallway) by constantly bouncing from the cladding (mirror-lined walls), a principle called total internal reflection. Because the cladding does not absorb any light from the core, the light wave can travel great distances.
However, some of the light signal degrades within the fiber, mostly due to impurities in the glass. The extent that the signal degrades depends on the purity of the glass and the wavelength of the transmitted light (for example, 850 nm = 60 to 75 percent/km; 1,300 nm = 50 to 60 percent/km; 1,550 nm is greater than 50 percent/km). Some premium optical fibers show much less signal degradation -- less than 10 percent/km at 1,550 nm.
1
The work done to pull the sled up to the hill is given by
where
F is the intensity of the force
d is the distance where the force is applied.
In our problem, the work done is
and the distance through which the force is applied is
, so we can calculate the average force by re-arranging the previous equation and by using these data:
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.
