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4 years ago

What is the speed between reflected ray and the incident ray

Physics

1 answer:

Korvikt [17]4 years ago

7 0

Answer:

The speed is the same as long as the reflection is regular.

Explanation:

This is because in regular reflection, the angle of incidence is equal to the angle of reflection in accordance with the second law of reflection.

Since speed of light depends on the angle of the light ray it makes with the reflecting surface, the speed is the same

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Using complete sentences, explain why graphs are so important to scientists.

morpeh [17]

Answer: It keeps them organized

Explanation: ... ew

5 0

3 years ago

Physics help please

zhuklara [117]

Answer: 37.981 m/s

Explanation:

This situation is related to projectile motion or parabolic motion, in which the travel of the ball has two components: x-component and y-component. Being their main equations as follows:

x-component:

$x=V_{o}cos\theta t$ (1)

Where:

$x=52 m$ is the point where the ball strikes ground horizontally

$V_{o}$ is the ball's initial speed

$\theta=0$ because we are told the ball is thrown horizontally

$t$ is the time since the ball is thrown until it hits the ground

y-component:

$y=y_{o}+V_{o}sin\theta t+\frac{gt^{2}}{2}$ (2)

Where:

$y_{o}=120m$ is the initial height of the ball

$y=0$ is the final height of the ball (when it finally hits the ground)

$g=-9.8m/s^{2}$ is the acceleration due gravity

Knowing this, let's start by finding $t$ from (2):

$0=y_{o}+V_{o}sin(0\°) t+\frac{gt^{2}}{2}$ (3)

$0=y_{o}+\frac{gt^{2}}{2}$

$t=\sqrt{\frac{-2 y_{o}}{g}}$ (4)

$t=\sqrt{\frac{-2 (120 m)}{-9.8m/s^{2}}}$ (5)

$t=4.948 s$ (6)

Then, we have to substitute (6) in (1):

$x=V_{o}cos(0\°) t$ (7)

And find $V_{o}$ :

$V_{o}=\frac{x}{t}$ (8)

$V_{o}=\frac{52 m}{4.948 s}$ (9)

$V_{o}=10.509 m/s$ (10)

On the other hand, since we are dealing with constant acceleration (due gravity) we can use the following equation to find the value of the ball's final velocity $V$ :

$V=V_{o} + gt$ (11)

$V=10.509 m/s + (-9.8 m/s^{2})(4.948 s)$ (12)

$V=-37.981 m/s$ (13) This is the ball's final velocity, and the negative sign indicates its direction is downwards.

However, we were asked to find the ball's final speed, which is the module of the ball's final vleocity vector. This module is always positive, hence the speed of the ball just before it strikes the ground is 37.981 m/s (positive).

5 0

3 years ago

What is the displacement of the particle in the time interval 5 seconds to 8 seconds?

Dmitriy789 [7]

The displacement is the vector with

magnitude
   distance between position at 5 sec and position at 8 sec
and direction
   direction from position at 5 sec to position at 8 sec .

The route followed during the time interval is irrelevant.

3 0

3 years ago

Read 2 more answers

Who was the first scientist to propose that an object could emit only certain amounts of energy?

Rzqust [24]

It was Niels Bohr who proposed it

3 0

3 years ago

Firecracker A is 300 m from you. Firecracker B is 600 m from you in the same direction. You see both explode at the same time. D

ss7ja [257]

Answer:

e see that the distances are different, the only way that the two beams of light approach simultaneously is that event 2 (farthest) occurs first than event 1

Explanation:

This is an ejercise in special relativity, where the speed of light is constant.

Let's carefully analyze the approach, we see the two events at the same time.

The closest event time is

c = (x₁-300) / t

t = (x₁-300) / c

The time for the other event is

t = (x₂- 600) / c

since they tell us that we see the events simultaneously, we can equalize

(x₁ -300) / c = (x₂ -600) / c

x₁ = x₂ - 300

We see that the distances are different, the only way that the two beams of light approach simultaneously is that event 2 (farthest) occurs first than event 1

3 0

3 years ago