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morpeh [17]
2 years ago
13

An electromagnetic wave of wavelength

Physics
1 answer:
Ivanshal [37]2 years ago
8 0

Answer:

4.01\cdot 10^{-7} m

Explanation:

When an electromagnetic wave passes through the interface between two mediums, it undergoes refraction, which means that it bents and its speed and its wavelength change.

In particular, the wavelength of an electromagnetic wave in a certain medium is related to the index of refraction of the medium by:

\lambda=\frac{\lambda_0}{n}

where

\lambda_0 is the wavelength in a vacuum (air is a good approximation of vacuum)

n is the refractive index of the medium

In this problem:

\lambda_0 = 5.89\cdot 10^{-7} m is the original wavelength of the wave

n = 1.47 is the index of refraction of corn oil

Therefore, the wavelength of the electromagnetic wave in corn oil is:

\lambda=\frac{5.89\cdot 10^{-7}}{1.47}=4.01\cdot 10^{-7} m

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Transform boundary

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Consider the two moving boxcars in Example 5. Car 1 has a mass of m1 = 65000 kg and a velocity of v01 = +0.80 m/s. Car 2 has a m
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Answer:

1.034m/s

Explanation:

We define the two moments to develop the problem. The first before the collision will be determined by the center of velocity mass, while the second by the momentum preservation. Our values are given by,

m_1 = 65000kg\\v_1 = 0.8m/s\\m_2 = 92000kg\\v_2 = 1.2m/s

<em>Part A)</em> We apply the center of mass for velocity in this case, the equation is given by,

V_{cm} = \frac{m_1v_1+m_2v_2}{m_1+m_2}

Substituting,

V_{cm} = \frac{(65000*0.8)+(92000*1.2)}{92000+65000}

V_{cm} = 1.034m/s

Part B)

For the Part B we need to apply conserving momentum equation, this formula is given by,

m_1v_1+m_2v_2 = (m_1+m_2)v_f

Where here v_f is the velocity after the collision.

v_f = \frac{m_1v_1+m_2v_2}{m_1+m_2}

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8 0
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mass of the bottle in each case is M = 0.250 kg

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K = \frac{1}{2}mv^2

K = \frac{1}{2}(0.250)(2)^2 = 0.5 J

2) when speed is 3 m/s

kinetic energy is given as

K = \frac{1}{2}mv^2

K = \frac{1}{2}(0.250)(3)^2 = 1.125 J

3) when speed is 4 m/s

kinetic energy is given as

K = \frac{1}{2}mv^2

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4) when speed is 5 m/s

kinetic energy is given as

K = \frac{1}{2}mv^2

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A test charge of 13 mC is at a point P where an external electric field is directed to the right and has a magnitude of 4 3 106
LenKa [72]

Answer:

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