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iren2701 [21]
3 years ago
13

A sound wave has a frequency of 425Hz. What is the period of this wave? a) 0.00235 b) 0.807 c) 425 d) 850

Physics
1 answer:
swat323 years ago
3 0

the answer is a) 0.00235 because 1/425=0.00235. hope I helped!

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Answer:

Helium

Explanation:

Helium is the least reactive element, since it is a noble gas with the smallest amount of valence rings.

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Which of the following best defines mass?
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A light beam travels at 1.94×108 in quartz. The wavelength of the light in quartz is 355 .Part AWhat is the index of refraction
Alja [10]

A) 1.55

The speed of light in a medium is given by:

v=\frac{c}{n}

where

c=3\cdot 10^8 m/s is the speed of light in a vacuum

n is the refractive index of the material

In this problem, the speed of light in quartz is

v=1.94\cdot 10^8 m/s

So we can re-arrange the previous formula to find n, the index of refraction of quartz:

n=\frac{c}{v}=\frac{3\cdot 10^8 m/s}{1.94\cdot 10^8 m/s}=1.55

B) 550.3 nm

The relationship between the wavelength of the light in air and in quartz is

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

where

\lambda is the wavelenght in quartz

\lambda_0 is the wavelength in air

n is the refractive index

For the light in this problem, we have

\lambda=355 nm\\n=1.55

Therefore, we can re-arrange the equation to find \lambda_0, the wavelength in air:

\lambda_0 = n\lambda=(1.55)(355 nm)=550.3 nm

4 0
3 years ago
How does the atmosphere’s interaction with different wavelengths of light help to protect life on earth?
marishachu [46]

Answer:

Ozone layer in the upper atmosphere filters most of the harmful radiations of shorter wavelength. It actually absorbs the hazardous radiations like ultraviolet, gamma rays, x- rays and most of all those having shorter wavelength then the visible light. That's how the earth's atmosphere protects life on earth. But unfortunately, climate change and global warming is causing the depletion of ozone layer which is causing skin related diseases and harming not only the human life but also the plants and animals.

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3 years ago
A charged particle enters a uniform magnetic field B with a velocity v at right angles to the field. It moves in a circle with p
alukav5142 [94]

A) d. 10T

When a charged particle moves at right angle to a uniform magnetic field, it experiences a force whose magnitude os given by

F=qvB

where q is the charge of the particle, v is the velocity, B is the strength of the magnetic field.

This force acts as a centripetal force, keeping the particle in a circular motion - so we can write

qvB = \frac{mv^2}{r}

which can be rewritten as

v=\frac{qB}{mr}

The velocity can be rewritten as the ratio between the lenght of the circumference and the period of revolution (T):

\frac{2\pi r}{T}=\frac{qB}{mr}

So, we get:

T=\frac{2\pi m r^2}{qB}

We see that this the period of revolution is directly proportional to the mass of the particle: therefore, if the second particle is 10 times as massive, then its period will be 10 times longer.

B) a. f/10

The frequency of revolution of a particle in uniform circular motion is

f=\frac{1}{T}

where

f is the frequency

T is the period

We see that the frequency is inversely proportional to the period. Therefore, if the period of the more massive particle is 10 times that of the smaller particle:

T' = 10 T

Then its frequency of revolution will be:

f'=\frac{1}{T'}=\frac{1}{(10T)}=\frac{f}{10}

6 0
3 years ago
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