1. Answer;
Velocity decreases as it rises
Explanation;
When you project an object upward and release it at its initial velocity, it is moving in the opposite direction of the force of gravity. Thus the initial velocity is negative. The velocity of the object is also negative on the way up but positive on the way down.
2. Answer;
Velocity is 0 m/s at its highest point
Explanation;
The maximum height of the projectile is when the projectile reaches zero vertical velocity. From this point the vertical component of the velocity vector will point downwards.
The object slows down as it moves upward until it reaches a maximum height, at which time the velocity is zero. Then the velocity increases as the object falls toward the ground.
3. Answer;
-Acceleration is constant -9.8m/s^2
Explanation;
-An object that is thrown vertically upwards decelerates under the earth's gravity. Its speed decreases until it attains a maximum height, where the velocity is zero. Then it is accelerated uniformly downwards under gravity.
4. Answer;
-Acceleration is constant -9.8m/s^2
Explanation;
When a ball is thrown upwards the acceleration due to gravity remains constant throughout. It comes to rest for a moment at the highest point of motion just before returning back to earth.
-The acceleration due to gravity is constant on the object though out its flight. So the acceleration of the projectile is equal to the acceleration due to gravity, 9.81 meters/second/second, from just after its thrown, through its highest point, and until just before it hits the ground.
Answer:
In physics, mass–energy equivalence is the principle that anything having mass has an equivalent amount of energy and vice ve
Answer: ssm An airplane with a speed of 97.5 m/s is climbing upward at an angle of 50.0° with respect to the horizontal. When the plane's altitude is 732 m, the pilot releases a package.
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}
