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4 years ago

Which type of electromagnetic waves make up the colors of a rainbow seen after a storm?

Physics

2 answers:

quester [9]4 years ago

7 0

The correct answer to the question is Visible light.

EXPLANATION:

A rainbow is visible to our naked eye. It contains all the seven constituent colors of a white light ranging from violet to red. Generally it is formed after a storm. It is due to the fact that different colors of light has different wavelength. When the white light falls on the water particles present in the atmosphere, all the colors deviate at different angles after their refraction through water particles, and the spectrum is obtained.

The infrared wave, X- rays and ultraviolet rays are not visible to our eye.

The visible light is that electromagnetic wave which helps us to view any object. It contains the seven colors just like a rainbow i.e ranging from violet to red.

All the colors present in a rainbow corresponds to visible region of the electromagnetic spectrum .

Hence, visible light is that electromagnetic wave which makes up the colors of a rainbow seen after a storm.

Step2247 [10]4 years ago

3 0

Visible light waves are the type of electromagnetic waves that make up the colors of the rainbow, because the rainbow is visible to us.

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Vladimir [108]

Answer:

$v_2=8.85\times 10^{-4}\ m/s$

Explanation:

We have,

Mass of a wrench, m₁ = 4 g = 0.004 kg

Speed of wrench, v₁ = -15 m/s

Mass of the Astronaut, m₂ = 70 kg

It is required to find Astronaut's velocity. Let it is v₂. Using the conservation of linear momentum to find it.

$m_1v_1=m_2v_2\\\\v_2=\dfrac{m_1v_1}{m_2}\\\\=\dfrac{0.004\times (-15)}{70}\\\\=-8.85\times 10^{-4}\ m/s$

So, his velocity is $8.85\times 10^{-4}\ m/s$ .

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3 years ago

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3 years ago

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4 years ago

A 10-g bullet moving horizontally with a speed of 1.8 km/s strikes and passes through a 5.0-kg block initially at rest on a hori

marshall27 [118]

Answer:

$K_2=6.4J$

Explanation:

According to the principle of conservation of momentum, we have:

$\Delta p=0\\p_i=p_f\\m_1v_1_i +m_2v_2_i=m_1v_1_f+m_2v_2_f$

Here 1 is for the bullet and 2 is for the block. Since the block is initially at rest $v_i_2=0$ . Solving for $v_2_f$ and replacing the given values:

$v_2_f=\frac{m_1(v_1_i-v_1_f)}{m_2}\\v_2_f=\frac{10*10^{-3}kg(1.8\frac{km}{s}-1\frac{km}{s})}{5kg}\\v_2_f=0.0016\frac{km}{s}*\frac{1000m}{1km}=1.6\frac{m}{s}$

The kinetic energy of the block is given by:

$K_2=\frac{m_2(v_f_2)^2}{2}\\K_2=\frac{5kg(1.6\frac{m}{s})^2}{2}\\K_2=6.4J$

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3 years ago