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Nesterboy [21]
3 years ago
9

When a DVD is played, a laser light hits the surface of the disk and then returns to the detector. An illustration of DVD with a

n arrow pointing toward a spot on the disk labeled detector. Which wave phenomenon allows the DVD player to work? absorption interference reflection refraction
Physics
1 answer:
Ierofanga [76]3 years ago
5 0

Answer:

Reflection

Explanation:

The wave phenomenon that allows the DVD player to work is REFLECTION.

Reflection is defined as the repropagation of incident light striking a plane surface.

The light ray striking the surface is incident ray while the repropagated light ray is the reflected ray. The DVD was able to play because the laser light that hits the surface of the disk was able to reflect back and returns to the detector. The detector senses the light and cause the DVD to play. If the laser light didn't reflect, assuming it was absorbed by the surface of the disk, the detector wouldn't have detected the light and the DVD wouldn't have played.

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A string attached to an airborne kite was maintained at an angle of 40.0 with the ground. if 120m of string was reeled in to ret
AleksAgata [21]
The hypotenuse is measured at 120 meters of string, and you need to solve for the leg of the triangle that is horizontal. The degree is 40, so use trigonometry to figure it out. 
Cosin (40) is equal to around .766 
Adjacent/Hypotenuse 
x/120 = cos40 
Answer: 91.92533. 
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5 0
3 years ago
Two horizontal metal plates, each 10.0 cm square, are aligned 1.00 cm apart with one above the other. They are given equal-magni
Ber [7]

Answer:

a) motion is PARABOLIC, b) positive particle is accelerated towards the negative plate,  c)  x = 6.19 10⁹ m

Explanation:

This exercise looks at the motion of a positively charged particle in an electric field.

a) Since the field is vertical the acceleration in this direction is

            F = m a

the electric force is

           F = q E

we substitute

          q E = m a

           a = qE / m

the mass of the particle is m = 2.00 10-16 kg

           a = 1.6 10⁻¹⁹ 2.02 10³ / 2.00 10⁻¹⁶ kg

           a = 1,616 m / s²

           

on the x-axis there are no relationships because there are no forces.

Since the particle has velocities in both axes, its motion is PARABOLIC,

b) the positive particle is accelerated towards the negative plate,

The field is descending, for which the event is down

c) where  hit the particle on the x-axis

they indicate that the particle leaves the center of the negative plate, for which we will fix our reference system at this point.

Let's find the components of the initial velocity.

           sin θ = v_{oy} / v

           cos θ = v₀ₓ / v

           v_{oy} = v₀ sin θ

           v₀ₓ = v₀ cos θ

           v_{oy) = 1.02 10⁵ sin 37 = 0.6139 10⁵ m / s

           v₀ₓ = 1.02 10⁵ cos 37 = 0.8146 10⁵ m / s

Let's find the time it takes to hit the negative plate

            y = y₀I + v_{oy} t + ½ a and t2

in this case the positions are y = y₀ = 0 and the accelerations

a = - 1,616m/s2,

we substitute

            0 = 0 + v_{oy} t - ½ a_y t²

            v_{oy}= ½ a_y t

            t = 2v_{oy} / a_y

let's calculate

           t = 2 0.6139 10⁵ / 1.616

           t = 7.597 10⁴s

in this time the particle travels a horizontal distance

           x = v₀ₓ t

           x = 0.8145 10⁵ 7.597 10⁴

           x = 6.19 10⁹ m

the particle falls off the plate

4 0
2 years ago
Describe the interaction with the sun and producers
bagirrra123 [75]

Answer: the sun's rays is one of the raw Materials recquired by plants to make food

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4 0
3 years ago
A football punter wants to kick the ball so that it is in the air for 4.5 s and lands 50 m from where it was kicked. Assume that
irakobra [83]

Answer:

(a) The angle of projection is 63 degree.

(b) The velocity of projection is 24.5 m/s.

Explanation:

Height, h = 1 m

horizontal distance, d = 50 m

time, t = 4.5 s

Let the initial velocity is u and the angle is A.

(a) Horizontal distance = horizontal velocity x time

50 = u cos A x 4.5

u cos A = 11.1 .....(1)

Use second equation of motion in vertical direction

h = u t + 0.5 gt^2\\\\- 1 = u sin A \times 4.5 - 0.5 \times 9.8\times 4.5^2\\\\u sin A = 21.8 ..... (2)

Divide (2) by (1)

tan A = 1.97

A = 63 degree

(b) Substitute the value of A in equation (2)

u x sin 63 = 21.8

u = 24.5 m/s

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Glider one and glider two collided. The data table above shows the momentum of each before and after the collision. Perform an a
k0ka [10]

I think there was momentum conserved

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