Answer:
Explanation:
Given:
- mass of bullet,
- initial velocity of bullet,
- displacement of the bullet in the target,
Here as given in the question the bullet penetrates the target by the given displacement of the bullet into it. During this process it faces deceleration and hence it comes to rest.
- so, final velocity of the bullet,
Now using the equation of motion:
where:
acceleration of the bullet
<u>Now the force of resistance offered by the target in stopping it:</u>
The ball has to push things out of its way in order to keep going.
For example:
-- billions of air molecules
-- hundreds of blades of grass .
-- To push something out of its way, the ball exerts a force against the
obstacle, and pushes it some distance.
-- A force acting through a distance is 'work', and that means 'energy'.
-- The ball has a certain amount of energy ... the amount you gave it
when you kicked it. It gives every air molecule and blade of grass a
little bit of energy, so the ball has less and less as it goes along.
-- Eventually, it has no more energy to use to push things out of the way.
No energy means no motion, so it stops.
-- If it were in space where there are no air molecules and no grass,
it would go on forever.
Answer:
Part a)
Part b)
Part c)
Explanation:
Part a)
As we know that frequency = 1 MHz
speed of electromagnetic wave is same as speed of light
So the wavelength is given as
Part b)
As we know the relation between electric field and magnetic field
Part c)
Intensity of wave is given as
Pressure is defined as ratio of intensity and speed
SOLUTION:
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?).
\begin{displaymath}
2t = \frac{\lambda_{film}}{2}.\end{displaymath}
The wavelength in the film is
\begin{displaymath}
\lambda_{film} = \frac{\lambda_{air}}{n} = \frac{565 nm}{1.38}\end{displaymath}
\begin{displaymath}
\lambda_{film} = 409 nm.\end{displaymath}
Hence, the thickness should be
\begin{displaymath}
t = \frac{\lambda_{film}}{4} = 102 nm .\end{displaymath}
