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Mariana [72]
3 years ago
5

Reflection occurs when a wave _____.

Physics
2 answers:
madam [21]3 years ago
8 0

Answer: The correct answer is " bounces off an object".

Explanation:

Reflection occurs when a wave bounces off an object.  

For example, when a ray of light incidents on the surface of a plane mirror. The light gets reflected from the plane mirror because of the silver polishing on the backside of the mirror.

There are two laws of reflection:

(1) The reflected ray, the incident ray and the normal to the surface of the mirror are all lie in the same plane.

(2) The angle of reflection is equal to the angle of incidence.

faust18 [17]3 years ago
3 0
When they bounce off a barrier
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Demuestre que la función de onda: y(x, t) = A e^−i(kx−ωt+φ) satisface la ecuación de onda lineal. (donde i =√−1).
Alekssandra [29.7K]

Answer:

La ecuación de onda lineal para una función f(x, t) se escribe como:

\frac{d^2f}{dt^2}  = c^2*\frac{d^2f}{dx^2}

Donde c es una constante que depende de la velocidad de propagación de la onda.

En este caso, tenemos:

f(x) = A*e^{-i*(k*x - w*t + \phi)}

Recordar que la derivada de la exponente es tal que:

k(x,y) = A*e^{c*x + n*y}\\dk/dx = c*A*e^{c*x + n*y}

Entonces las derivadas de f van a ser:

\frac{df}{dt} =  (i*w)*f(x)

\frac{d^2f}{dt^2} = (i*w)^2*f(x) = -(w)^2*f(x)

\frac{df}{dx} = (-i*k)*f(x)

\frac{d^2f}{dx^2} = (-i*k)^2*f(x) = -k^2*f(x)

Entonces podemos reescribir la ecuación de onda como:

\frac{d^2f}{dt^2}  = c^2*\frac{d^2f}{dx^2}

-(w)^2*f(x) = c^2*(-k^2*f(x))

Ahora podemos simplemente definir c^2 = (w/k)^2 y vemos que la ecuación de onda se cumple.

5 0
3 years ago
During a safety crash, a car impacts a solid concrete barrier and and comes to rest in 0.10s. In a second test, an identical car
kirill [66]

Answer:

The correct answer is C

Explanation:

Change is momentum can be described as the change in the product of mass and velocity of a body. Every moving object as a momentum and the higher the momentum of this object, the harder it is to stop. Impulse (a force), which is sometimes used to describe change in momentum can be described as the product as force multiplied by time.

From the description above, it can be deduced that an increase in impulse can lead to a greater change in momentum. And an increase in impulse can be brought about by an increase in the time it takes a body to be brought to rest after collision. And since the car that hit the water barrels was brought to rest at a longer time, it has a greater change in momentum

8 0
3 years ago
How must of the roller coaster's kinetic and/or potential energy is in each of the points. For each point, pick one option. (Sor
labwork [276]

Answer

  1. Point A-
  2. Point B-
  3. Point C- 100 Potential, 0 Kinetic

3 0
4 years ago
What's the way to building a time machine and I will reward you beautifully.DENY PHYSICS YOUNG JEDI
quester [9]
Easy brother. You just need some luck.

Inside every black hole is a time machine that allows you to turn back time or go forward and teleport to any empty space in the universe. However to do so you need to get inside the black hole without dieing.

To do so you must make the hardest and lightest box you can make, fill it with a very dense kind of air that will allow you to breathe but not move. Then seal yourself In the box.
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4 0
3 years ago
I really need help with the graphs
nata0808 [166]

Answer:

t = 2s

Explanation:

When you're looking for instantaneous portions of a graph, of any sort really, it means you're observing a rate at a single point in time [or possibly some other variable]. It's sorta like a snapshot of a rate as opposed to an average rate over an interval. After choosing this rate we'll typically draw a straight, tangent line through it to indicate it's slope. (Tangent lines are just lines that only touch a single point on a graph or shape.)

Another thing to take note of are the values of the graph's major axes. The "y-axis" corresponds to velocity in meters per second, while the "x-axis" corresponds to time in seconds. Normally when relating the two we put "y" over the "x" and say that at any point there are "y[units]" per "x[units]". Though with instantaneous rates, we say the value of "x" is "1"; for reasons I can try to further explain later if you'd like.

With the above information in mind we can turn our attention to your graph. You're told to find the point on this graph where the instantaneous rate of acceleration is -2 m/s². The only place where the graph reflects an instantaneous rate of -2m/s² is at t = 2s. At t = 2, the rate comes out to (2[m/s]/1s), which simplifies to 2m/s². If you then draw the tangent line through the point, you'll find that the line is decreasing (going down from left to right) which means that the instantaneous rate is negative.

So at t = 2s, we have an instantaneous acceleration of -2m/s².

3 0
3 years ago
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