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3 years ago

All of the visible color light waves together make up

Physics

1 answer:

Dafna11 [192]3 years ago

8 0

C. white light

Explanation:

All the visible color light waves are seen together and makes up a white light.

Visible light is made up of different colors and when they join together they are seen as the white light around us.

During rainbow formation, we see the different colors of light that makes up a white light.
when white light passes through a glass prism, the velocity and wavelength changes and we can see the different components of light.

learn more:

electromagnetic waves brainly.com/question/12450147

#learnwithBrainly

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A pendulum oscillates 12 times in 4 seconds. what is the length of the pendulum?

seraphim [82]

Answer:

L = 2.8 cm

Explanation:

Period T = 4 / 12 = 1/3 s

T = 2π√(L/g)

L = (T/2π)²g

L = ((1/3)/2π)²9.8 = 0.02758... ≈ 2.8 cm

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3 years ago

Need help will mark you the Brainliest

Hatshy [7]

The last one is correct (D)

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3 years ago

For her birthday, amy received flowers that had a dull red appearance. compared with the entire range of visible light waves, th

Cerrena [4.2K]

The flowers reflect relatively low frequency and low amplitude light waves.
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3 years ago

A potential difference of 24 V is applied to a 150-ohm resistor. How much current flows through the resistor?

Lady_Fox [76]

Given :- A resistor of 150 ohm, hence Resistance (R) = 150 ohm

Potential Difference (v) = 24 V

Current (I) = ?

V = IR

24 = I × 150

I = 24/150

I = 0.16 ampere

hope it helps!

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3 years ago

A 15x10^-6c charge is placed at the origin and a 9x10^-6C charge is placed on the x-axis at x=1.00m. where, on the x-axis is the

Harlamova29_29 [7]

The Electric field is zero at a distance 2.492 cm from the origin.

Let A be point where the charge $15\times10^-6$ C is placed which is the origin.

Let B be the point where the charge $9\times 10^-6$ C is placed. Given that B is at a distance 1 cm from the origin.

Both the charges are positive. But charge at origin is greater than that of B. So we can conclude that the point on the x-axis where the electric field = 0 is after B on x - axis.

i.e., at distance 'x' from B.

Using Coulomb's law, $\frac{kQ_A^2}{d_A^2} = \frac{kQ_B^2}{d_B^2}$ ,