Answer:
The answer is 24
Explanation:
Its made up of 6 carbon atoms
6 oxygen atoms
12 hydrogen atoms
Since the two charts after the collision stick together, we are dealing with a perfectly inelastic collision.
First, we need to find the speed of the two charts after the collision. In order to do so, we consider the conservation of momentum:
m₁·v₁ + m₂·v₂ = (m₁ + m₂)·v
We can solve for v, considering also that v₂=0
v = m₁·v₁ / (m₁ + m<span>₂)
= 0.31 </span>· 0.90 / (0.31 + 0.50)
= 0.34 m/s
The kinetic energy lost (which is transformed into bounding energy between the two charts) will be the difference between the total kinetic energy before the collision and after the collision:
ΔE = E₁ - E₂ = 1/2·m₁·v₁ - 1/2·(m₁ + m₂)·v
= 1/2(0.31)(0.90) - 1/2(0.81)(0.34)
= 0.1395 - 0.1377
= 0.0018J
Hence, the correct answer is ΔE = <span>0.0018J</span>
Answer:
Can't tell if this is a real question or not but....
Reality shifting is based around this idea, but instead of letting it happen freely, the most precise details of your experience are planned by you! It's like incorporating a daydream into your subconscious. Reality shifting is a complex process, but anyone is able to do it.
Explanation:
Answer:
I know the answer
Explanation:
We want to choose the film thickness such that destructive interference occurs between the light reflected from the air-film interface (call it wave 1) and from the film-lens interface (call it wave 2). For destructive interference to occur, the phase difference between the two waves must be an odd multiple of half-wavelengths.
You can think of the phases of the two waves as second hands on a clock; as the light travels, the hands tick-tock around the clock. Consider the clocks on the two waves in question. As both waves travel to the air-film interface, their clocks both tick-tock the same time-no phase difference. When wave 1 is reflected from the air-film boundary, its clock is set forward 30 seconds; i.e., if the hand was pointing toward 12, it's now pointing toward 6. It's set forward because the index of refraction of air is smaller than that of the film.
Now wave 1 pauses while wave two goes into and out of the film. The clock on wave 2 continues to tick as it travels in the film-tick, tock, tick, tock.... Clock 2 is set forward 30 seconds when it hits the film-lens interface because the index of refraction of the film is smaller than that of the lens. Then as it travels back through the film, its clock still continues ticking. When wave 2 gets back to the air-film interface, the two waves continue side by side, both their clocks ticking; there is no change in phase as they continue on their merry way.
So, to recap, since both clocks were shifted forward at the two different interfaces, there was no net phase shift due to reflection. There was also no phase shift as the waves travelled into and out from the air-film interface. The only phase shift occured as clock 2 ticked inside the film.
Call the thickness of the film t. Then the total distance travelled by wave 2 inside the film is 2t, if we assume the light entered pretty much normal to the interface. This total distance should equal to half the wavelength of the light in the film (for the minimum condition; it could also be 3/2, 5/2, etc., but that wouldn't be the minimum thickness) since the hand of the clock makes one revolution for each distance of one wavelength the wave travels (right?).
Impulse is the equivalent to the force as the change in momentum, such that F*t = mv - mu
Since mass and speed is given, use the above right hand side formula mass, m = 1.7 x 10^27 kg initial velocity, u = 0.991 x (3 x 10^8) = 2.973 x 10^8 final velocity, v = 0.994 x (3 x 10^8) = 2.982 x 10^8
So the computation is:
Impulse = mv - mu
= [(1.7 x 10^27) x (2.982 x 10^8)] - [(1.7 x 10^27) x (2.973 x 10^8)] = (5.0694 x 10^35) - (5.0694 x 10^35) = 0