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4 years ago

What is it called when a surface takes light in without reflecting it

Physics

2 answers:

andrew-mc [135]4 years ago

7 0

Reflection from such a rough surface is called diffuse reflection and appears matte

Travka [436]4 years ago

5 0

Answer: Specular reflection

Explanation:

The law of reflection says that when a ray of light hits a surface, it bounces in a certain way, like a tennis ball thrown against a wall. ... Mirrors, however, don't scatter light in this way. With a smooth surface, light reflects without disturbing the incoming image. This is called specular reflection.

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An electron enters a region with a speed of 5×10^6m/s and is slowed down at the rate of 1.25×10^-4m/s². How far does the electro

Mashutka [201]

1) The distance travelled by the electron is $1\cdot 10^{17} m$

2) The time taken is $4.0\cdot 10^{10}s$

Explanation:

The electron in this problem is moving by uniformly accelerated motion (constant acceleration), so we can use the following suvat equation

$v^2-u^2=2as$

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance travelled

For the electron in this problem,

$u=5\cdot 10^6 m/s$ is the initial velocity

v = 0 is the final velocity (it comes to a stop)

$a=-1.25\cdot 10^{-4} m/s^2$ is the acceleration

Solving for s, we find the distance travelled:

$s=\frac{v^2-u^2}{2a}=\frac{0-(5\cdot 10^6)^2}{2(-1.25\cdot 10^{-4})}=1\cdot 10^{17} m$

The total time taken for the electron in its motion can also be found by using another suvat equation:

$v=u+at$

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time taken

Here we have

$u=5\cdot 10^6 m/s$

v = 0

$a=-1.25\cdot 10^{-4} m/s^2$

And solving for t, we find the time taken:

$t=\frac{v-u}{a}=\frac{0-5\cdot 10^6}{-1.25\cdot 10^{-4}}=4.0\cdot 10^{10}s$

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

7 0

3 years ago

How do I answer this:

11111nata11111 [884]

The following are the answers to the question presented:

a. magnitude of the radial acceleration = 1.25m/s² inwardly directed

b. tangential acceleration = 0.400m/s²

c. total acceleration = 72.25 degrees

I am hoping that these answers have satisfied your queries and it will be able to help you in your endeavors, and if you would like, feel free to ask another question.

8 0

3 years ago

Read 2 more answers

Which one of the following formulas is used to find the voltage of a circuit?

Sedbober [7]

D , since Voltage is one joule per coulomb

8 0

3 years ago

If a dolphin swims at 15 m/s for 30 seconds, how far would the dolphin swim?

Olegator [25]

It would swim 30*15 metres, which is 450 metres.

7 0

3 years ago

An electron emitted from a filament is travelling at 1.5 x 105 m/s when it enters an acceleration of an electron gun in a televi

Crank

Answer:

The acceleration of the electron is 1.457 x 10¹⁵ m/s².

Explanation:

Given;

initial velocity of the emitted electron, u = 1.5 x 10⁵ m/s

distance traveled by the electron, d = 0.01 m

final velocity of the electron, v = 5.4 x 10⁶ m/s

The acceleration of the electron is calculated as;

v² = u² + 2ad

(5.4 x 10⁶)² = (1.5 x 10⁵)² + (2 x 0.01)a

(2 x 0.01)a = (5.4 x 10⁶)² - (1.5 x 10⁵)²

(2 x 0.01)a = 2.91375 x 10¹³

$a = \frac{2.91375 \ \times \ 10^{13}}{2 \ \times \ 0.01} \\\\a = 1.457 \ \times \ 10^{15} \ m/s^2$

Therefore, the acceleration of the electron is 1.457 x 10¹⁵ m/s².

7 0

3 years ago