2 years ago

The “rainbow” surface of a CD or DVD is the result of a transmission grating.

Physics

1 answer:

Gelneren [198K]2 years ago

6 0

Answer:

I believe a CD or DVD rainbow is actually reflection grating, but...

So I believe this is false.

Though it is possible to make it transmissive.

Explanation:

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Electromagnetic waves differ only in their _____, their energy, and their frequency.

Tanya [424]

Answer:

Wavelength

Explanation:

Electromagnetic Waves have different wavelengths.

They only differ from each other in wavelength. Wavelength is the distance between one wave crest to the next.

6 0

3 years ago

In the figure a block slides along a track from one level to a higher level, by moving through an

Scrat [10]

Answer:

(B) 1.17 m.

Explanation:

At the highest point, the conservation of energy equation would becomes;

K.E (initial ) = K.E + P.E;

½ m(V)^2 = ½ m(V)^2 + mgh

(m*6^2)/2 = (m * 9.81 * 1.1) + (m*V^2)/2

36/2 = (9.81 * 1.1) + v^2/2

Solving the equation;

v=√14.44 m/s

= 3.8 m/s

a = −0.6g

= -0.6*9.81

= −5.88 m/s^2

Using equation of motion,

V^2 = U^2 + 2aS

S = V^2/2a

= 14.44/12

= 1.19 m.

7 0

3 years ago

Graph the following data tables on different graphs.

Anna [14]

Answer:

Sjjsjsjsjsjsjsjwjwjw

8 0

2 years ago

Kyle is wondering what he is getting for his birthday. He decides to shake the box that contains his present. Kyle did not reali

stiks02 [169]

Answer: Asking questions

Explanation:

Scientists always start with their question before

Observing anything

8 0

2 years ago

A 5.0-μC charge is placed at the 0 cm mark of a meter stick and a -4.0 μC charge is placed at the 50 cm mark. At what point on a

Maksim231197 [3]

Answer:

The distance from charge 5 μ C = 26.45 cm and the distance from - 4 μ C is 23.55 cm.

Explanation:

Given that

q₁ = 5 μ C

q₂ = - 4 μ C

The distance between charges = 50 cm

d= 50 cm

Lets take at distance x from the charge μ C ,the electrical field is zero.

That is why the distance from the charge - 4 μ C = 50 - x cm

We know that ,electric field is given as

$E=K\dfrac{q}{r^2}$

$K\dfrac{5\ \mu}{x^2}=K\dfrac{4\mu }{(50-x)^2}\\\\\dfrac{5}{x^2}=\dfrac{4 }{(50-x)^2}\\\\\\5(50-x)^2=4x^2\\(50-x)^2=0.8x^2\\\\50-x =0.89x\\\ x=\dfrac{50}{1.89}\ cm\\\\\\x=26.45\ cm\\$

Therefore the distance from charge 5 μ C = 26.45 cm and the distance from - 4 μ C is 23.55 cm.

3 0

2 years ago