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Olegator [25]
2 years ago
7

How does reflection differ from refraction and diffraction?

Physics
1 answer:
shtirl [24]2 years ago
7 0

Answer: Reflection is the only process in which the wave does not continue moving forward.

Explanation:

Reflection is a process in which the direction of the wave changes when it is exposed to a bounce off barrier. Refraction can be defined as the change in the direction of the wave when the wave passes through one medium to another. Diffraction is a process in which the direction of the wave changes when the wave passes through a particular opening near the barrier.

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When waves interact do they move through each other or bounce off each other? WILL MARK BRAINLIEST!!!
4vir4ik [10]
Will the waves bounce off each other upon meeting or will the two waves pass through each water? Waves interference occurs when two waves meet while traveling along the same medium
4 0
3 years ago
The velocity of the transverse waves produced by an earthquake is 5.08 km/s, while that of the longitudinal waves is 8.3312 km/s
ASHA 777 [7]

Answer:

727.67 km

Explanation:

Sine they have Same distance D

distance = speed * time

D = 5.08t

D = 8.3312(t+55.9)

so

5.08t = 8.3312(t+55.9) t in

3.2512t = 465.71

t = 143.2s

Subtitute t

D=5.08 t

= 5.08 × 143.2

= 727.67km

7 0
2 years ago
____is the distance traveled during a specific unit of time
qaws [65]
I think speed is the answer
6 0
3 years ago
What fills the void between stars and galaxies.
Lana71 [14]
Answer: The voids between stars in our galaxy can be filled with tenuous clouds of gas and other molecules. ... That material gets "ripped away" from the galaxies by the force of gravity, and often enough it collides with other material.

HOPE IT HELPED:) HAVE A NICE DAY
6 0
3 years ago
You're driving down the highway late one night at 20 m/s when a deer steps onto the road 38 m in front of you. Your reaction tim
Bingel [31]

Answer:

given,

speed of the car = 20 m/s

final speed of car = 0 m/s

distance between car and the deer = 38 m

reaction time, t = 0.5 s

deceleration of the car = 10 m/s².

a) distance between deer and car

  distance travel in the reaction time

   d₁ = v x t

   d₁ = 20 x 0.5 = 10 m

   distance travel after you apply brake

   using equation of motion

   v² = u² + 2 a s

   0 = 20² - 2 x 10 x s

    s =  20 m

total distance traveled by the car

D = d₁ + d₂

D = 20 + 10 = 30 m

  distance between car and the deer = 38 m - 30 m

                                                              = 8 m

b) now, maximum speed car.

   distance travel in reaction time

    d₁ = s x t

    d₁ = 0.5 V

distance left between them

   d₂ = 38 - d₁

   d₂ = 38 - 0.5 V

   distance travel after you apply brake

   using equation of motion

    v² = u² + 2 a d₂

    0 = (V)² - 2 x 10 x (38 - 0.5 V)

     V² + 10 V - 760 = 0

now, solving the quadratic equation

  x = \dfrac{-b\pm \sqrt{b^2-4ac}}{2a}

  V = \dfrac{-10\pm \sqrt{10^2-4(1)(-760)}}{2(1)}

         V = 23.01 , -33.01

rejecting the negative term.

hence, maximum speed of the car could be V = 23.01 m/s

 

7 0
3 years ago
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