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Vaselesa [24]
4 years ago
9

A band of blue light has a velocity of 3.0 x 108

Physics
1 answer:
mr_godi [17]4 years ago
4 0

Answer: 6.4 x 10^14Hz

Explanation:

velocity of blue light (V) = 3.0 x 10^8m/sec

wavelength (λ) = 465 nanometers

(4.65 x 10?m)

Since 1nanometer = 1 x 10^-9 meter

465 nanometer = 4.65 x 10^-7 meters

frequency (F) = ?

Recall that the frequency of a wave is the number of cycles the wave complete in one second, and its unit is Hertz.

So, apply the formula V = F λ

3.0 x 10^8m/sec = F x 4.65 x 10^-7 meters

F = (3.0 x 10^8m/sec / 4.65 x 10^-7 meters)

F = 6.4 x 10^14Hz

Thus, the frequency of the blue light is 6.4 x 10^14Hz

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4 years ago
A violin string is 45.0 cm long and has a mass of 0.242 g. When tightened on the neck of the violin, the distance between the pi
stiks02 [169]

Answer:

The tension is 75.22 Newtons

Explanation:

The velocity of a wave on a rope is:

v=\sqrt{\frac{TL}{M}} (1)

With T the tension, L the length of the string and M its mass.

Another more general expression for the velocity of a wave is the product of the wavelength (λ) and the frequency (f) of the wave:

v= \lambda f (2)

We can equate expression (1) and (2):

\sqrt{\frac{TL}{M}}=\lambda f

Solving for T

T= \frac{M(\lambda f)^2}{L} (3)

For this expression we already know M, f, and L. And indirectly we already know λ too. On a string fixed at its extremes we have standing waves ant the equation of the wavelength in function the number of the harmonic N_{harmonic} is:

\lambda_{harmonic}=\frac{2l}{N_{harmonic}}

It's is important to note that in our case L the length of the string is different from l the distance between the pin and fret to produce a Concert A, so for the first harmonic:

\lambda_{1}=\frac{2(0.425m)}{1}=0.85 m

We can now find T on (3) using all the values we have:

T= \frac{2.42\times10^{-3}(0.85* 440)^2}{0.45}

T=75.22 N

3 0
3 years ago
Through what voltage must an αα-particle, with its charge of +2e+2e, be accelerated so that it has just enough energy to reach a
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Answer:5101.35v

Explanation:

Radius of gold nucleus=7.3×10-15m and a charge of +79e

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The nucleus is considered as the point charge where the potential energy between the charges are

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Where r is distance between the charges and the nucleus

r=R+d

V=U/q

U= 1/(4×3.142×Eo)×Q/r

V= 1/(4×3.142×Eo)×Q/(r+d)

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V= 9.0×10^9 ×(1.264×10^-17)/(2.23×10^-14)

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Answer:

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makkiz [27]
I think it is B. 100 W
7 0
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