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Ludmilka [50]
3 years ago
15

The wavelength of violet light is about 425 nm (1 nanometer = 1 × 10−9 m). what are the frequency and period of the light waves?

Physics
2 answers:
BigorU [14]3 years ago
4 0

1. Frequency: 7.06\cdot 10^{14} Hz

The frequency of a light wave is given by:

f=\frac{c}{\lambda}

where

c=3\cdot 10^{-8} m/s is the speed of light

\lambda is the wavelength of the wave

In this problem, we have light with wavelength

\lambda=425 nm=425\cdot 10^{-9} m

Substituting into the equation, we find the frequency:

f=\frac{c}{\lambda}=\frac{3\cdot 10^{-8} m/s}{425\cdot 10^{-9} m}=7.06\cdot 10^{14} Hz


2. Period: 1.42 \cdot 10^{-15}s

The period of a wave is equal to the reciprocal of the frequency:

T=\frac{1}{f}

The frequency of this light wave is 7.06\cdot 10^{14} Hz (found in the previous exercise), so the period is:

T=\frac{1}{f}=\frac{1}{7.06\cdot 10^{14} Hz}=1.42\cdot 10^{-15} s


marusya05 [52]3 years ago
4 0

ANSWER:

frequency = 7.058*10^14 Hz

time period = 1.41*10^-15 s

EXPLANATION:

by formula

f = c / λ

c is constant which is = 3 * 10^8

λ is given = 425 * 10^-9 m

f= 3* 10^8 / 425 * 10^-9

f = 7.058*10^14 Hz

T= 1/f

T = 1 / 705882.3529

T= 1.41*10^ -15 s

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3 years ago
A certain superconducting magnet in the form of a solenoid of length 0.300 m can generate a magnetic field of 8.90 T in its core
Ivan

Answer:

The number of turns in the solenoid is 22366.

Explanation:

The number of turns in the solenoid can be found using the following equation:

B = \mu_{0} I\frac{N}{L}

Where:

B: is the magnetic field = 8.90 T

L: is the solenoid's length = 0.300 m

N: is the number of turns =?

I: is the current = 95 A

μ₀: is the magnetic constant = 4π×10⁻⁷ H/m

By solving equation (1) for N we have:

N = \frac{BL}{\mu_{0} I} = \frac{8.90 T*0.300 m}{4\pi \cdot 10^{-7} H/m*95 A} = 22366 turns

Therefore, the number of turns in the solenoid is 22366.

I hope it helps you!

8 0
3 years ago
The wavelength of red helium-neon laser light in air is 632.8 nm.(a) What is its frequency? Hz(b) What is its wavelength in glas
vekshin1
<h2>Answers:</h2>

The speed of a wave is given by:

v=f.\lambda  (1)

Where f is the frequency and  \lambda the wavelength.

In the case of light, its speed is:

c=f.\lambda (2)

On the other hand, the described situation is known as Refraction,   a phenomenon in which the light changes its direction when passing through a medium with a refractive index different from the other medium.  

In this context, the Refractive index n is a number that describes how fast light propagates through a medium or material, and is defined as the relation between the speed of light in vacuum (c=3(10)^{8}m/s) and the speed of light v in the second medium:

n=\frac{c}{v} (3)

In addition, as the light changes its direction, its wavelength changes as well:

n=\frac{\lambda_{air}}{\lambda_{glass}} (4)

Knowing this, let's begin with the answers:

<h2>a) Frequency</h2>

From equation (2) we can find f:

f=\frac{c}{\lambda}  (5)

Knowing that 1nm=(10)^{-9}m:

f=\frac{3(10)^{8}m/s}{632.8(10)^{-9}m}  

f=4.74(10)^{14}Hz}     (6)   >>>Frequency of the helium-neon laser light

<h2>b) Wavelength in glass</h2>

We already know the wavelength of the light in air \lambda_{air} and the index of refraction of the glass.

So, we only have to find the wavelength in glass \lambda_{glass} from equation (4):

\lambda_{glass}=\frac{\lambda_{air}}{n}

\lambda_{glass}=\frac{632.8(10)^{-9}m}{1.48}

\lambda_{glass}=427(10)^{-9}m=427nm   (7)   >>>Wavelength of the helium-neon laser light in glass

<h2>c) Speed in glass</h2>

From equation (3) we can find the speed vof this light in glass:

v=\frac{c}{n}

v=\frac{3(10)^{8}m/s}{1.48}

v=2.027(10)^{8}m/s   (8)  >>>Speed of the helium-neon laser light in glass

7 0
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Answer:

A vacuum

Explanation:

Sound waves are examples of mechanical waves. Mechanical waves are waves which are transmitted through the vibrations of the particles in a medium.

For example, sound waves in air consist of oscillations of the air particles, which vibrate back and forth (longitudinal wave) along the direction of propagation of the wave itself.

Given this definition of mechanical wave, we see that such a wave cannot propagate if there is no medium, because there are no particles that would oscillate. Therefore, among the choices given, the following one:

a vacuum

represent the only situation in which a sound wave cannot propagate through: in fact, there are no particles in a vacuum, so the oscillations cannot occur. In all other cases, instead, sound waves can propagate.

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