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igomit [66]
3 years ago
5

Describe the difference between red light and blue light.

Physics
1 answer:
Anettt [7]3 years ago
8 0

Answer:

Red light has longer wavelength, while blue light has higher frequency

Explanation:

Red light and blue light corresponds to two different portion of the visible part of the electromagnetic spectrum.

The two colors have different wavelengths - more precisely:

Blue: 450 - 500 nm

Red: 620 - 750 nm

So, we see that red light has a longer wavelength than blue light.

Moreover, the frequency of an electromagnetic wave, f, is inversely proportional to the wavelength, \lambda:

f=\frac{c}{\lambda}

where c is the speed of light. This means that blue light, which has a shorter wavelength, has a higher frequency than red light.

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16x to the power of 6
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2. Suppose that a parallel-plate capacitor has circular plates with radius R = 30 mm and a plate separation of d = 5.0 mm. Suppo
seropon [69]

Answer:

The maximum value of the induced magnetic field is 2.901\times10^{-13}\ T.

Explanation:

Given that,

Radius of plate = 30 mm

Separation = 5.0 mm

Frequency = 60 Hz

Suppose the maximum potential difference is 100 V and r= 130 mm.

We need to calculate the angular frequency

Using formula of angular frequency

\omega=2\pi f

Put the value into the formula

\omega=2\times\pi\times60

\omega=376.9\ rad/s

When r>R, the magnetic field is inversely proportional to the r.

We need to calculate the maximum value of the induced magnetic field that occurs at r = R

Using formula of magnetic filed

B_{max}=\dfrac{\mu_{0}\epsilon_{0}R^2\timesV_{max}\times\omega}{2rd}

Where, R = radius of plate

d = plate separation

V = voltage

Put the value into the formula

B_{max}=\dfrac{4\pi\times10^{-7}\times8.85\times10^{-12}\times(30\times10^{-3})^2\times100\times376.9}{2\times130\times10^{-3}\times5.0\times10^{-3}}

B_{max}=2.901\times10^{-13}\ T

Hence, The maximum value of the induced magnetic field is 2.901\times10^{-13}\ T.

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Which object will sink in freshwater, which has a density of 1.0 g/cm3?
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The distance between the two slits is 40.11 μm.

Explanation:

Given that,

Frequency f= 6.37\times10^{14}\ Hz

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Position of the third order y =3.10 cm

We need to calculate the wavelength

Using formula of wavelength

\lambda=\dfrac{c}{f}

where, c = speed of light

f = frequency

Put the value into the formula

\lambda=\dfrac{3\times10^{8}}{6.37\times10^{14}}

\lambda=471\ nm

We need to calculate the distance between the two slits

m\times \lambda=d\sin\theta

d =\dfrac{m\times\lambda}{\sin\theta}

Where, m = number of fringe

d = distance between the two slits

Here, \sin\theta =\dfrac{y}{l}

Put the value into the formula

d=\dfrac{3\times471\times10^{-9}\times88.0\times10^{-2}}{3.10\times10^{-2}}

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