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 correct response that will be used to describe this particular element would be the third option, since all of the other options are incorrect and apply to different elements in their groups. The element is a metal and will react with a non metal.
Answer:

Explanation:
According to Pascal's Law, the pressure transmitted from input pedal to the output plunger must be same:

where,
F₁ = Load lifted by output plunger = 2100 N
F₂ = Force applied on input piston = 44 N
r₁ = radius of output plunger
r₂ = radius of input piston
Therefore,

Answer:
Explanation:
I got everything but i. Don't know why but it's eluding me. So let's do everything but that.
a. PE = mgh so
PE = (2.5)(98)(14) and
PE = 340 J
b.
so
and
KE = 250 J
c. TE = KE + PE so
TE = 340 + 250 and
TE = 590 J
d. PE at 8.7 m:
PE = (2.5)(9.8)(8.7) and
PE = 210 J
e. The KE at the same height:
TE = KE + PE and
590 = KE + 210 so
KE = 380 J
f. The velocity at that height:
and
so
v = 17 m/s
g. The velocity at a height of 11.6 m (these get a bit more involed as we move forward!). First we need to find the PE at that height and then use it in the TE equation to solve for KE, then use the value for KE in the KE equation to solve for velocity:
590 = KE + PE and
PE = (2.5)(9.8)(11.6) so
PE = 280 then
590 = KE + 280 so
KE = 310 then
and
so
v = 16 m/s
h. This one is a one-dimensional problem not using the TE. This one uses parabolic motion equations. We know that the initial velocity of this object was 0 since it started from the launcher. That allows us to find the time at which the object was at a velocity of 26 m/s. Let's do that first:
and
26 = 0 + 9.8t and
26 = 9.8t so the time at 26 m/s is
t = 2.7 seconds. Now we use that in the equation for displacement:
Δx =
and filling in the time the object was at 26 m/s:
Δx = 0t +
so
Δx = 36 m
i. ??? In order to find the velocity at which the object hits the ground we would need to know the initial height so we could find the time it takes to hit the ground, and then from there, sub all that in to find final velocity. In my estimations, we have 2 unknowns and I can't seem to see my way around that connundrum.
Red shirt appears red because:
a. The shirt reflects red light