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3 years ago

If the speed of a wave is 1500m/sec and its frequency is 200 Hz, what is its wavelength

Physics

1 answer:

Ray Of Light [21]3 years ago

4 0

Answer:

The wavelength of wave is 7.5 meter.

Given:

Speed of wave = 1500 $\frac{m}{s}$

Frequency of wave = 200 Hz

To find:

Wavelength of wave = ?

Formula used:

$\lambda = \frac{v}{n}$

Where $\lambda$ = wavelength of the wave

v = speed of wave

n = frequency of wave

Solution:

Wavelength of wave is given by,

$\lambda = \frac{v}{n}$

Where $\lambda$ = wavelength of the wave

v = speed of wave

n = frequency of wave

$\lambda = \frac{1500}{200}$

$\lambda$ = 7.5 m

The wavelength of wave is 7.5 meter.

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In a physics lab, light with a wavelength of 570 travels in air from a laser to a photocell in a time of 16.5 . When a slab of g

aliina [53]

Answer:

Wavelength is calculated as 213.9 nm

Solution:

As per the question:

Wavelength of light = 570 nm

Time, t = 16.5 ns

Thickness of glass slab, d = 0.865 ns

Time taken to travel from laser to the photocell, t' = 21.3

Speed of light in vacuum, c = $3\times 10^{8}\ m/s$

Now,

To calculate the wavelength of light inside the glass:

After the insertion of the glass slab into the beam, the extra time taken by light to cover a thickness t = 0.865 m is:

t' - t = 21.3 - 16.5 = 4.8 ns

Thus

$\frac{d}{\frac{c}{n}} - \frac{d}{v} = 4.8\times 10^{- 9}$

$\frac{0.8656}{\frac{c}{n}} - \frac{0.865}{v} = 4.8\times 10^{- 9}$

where

n = refractive index of the medium

v = speed of light in medium

$\frac{0.8656}{\frac{c}{n}} - \frac{0.865}{v} = 4.8\times 10^{- 9}$

$n = \frac{4.8\times 10^{- 9}\times 3.00\times 10^{8}}{0.865} + 1$

n = 2.66

Now,

The wavelength in the glass:

$\lambda' = \frac{\lambda }{n}$

$\lambda' = \frac{570}{2.66} = 213.9\ nm$

5 0

3 years ago

a body of radius R and mass m is rolling horizontally without slipping with speed v. it then rolls us a hill to a maximum height

ki77a [65]

Answer:

mR²/2

Explanation:

Here is the complete question

An object of radius′

R′ and mass ′

M′ is rolling horizontally without slipping with speed ′

V′

. It then rolls up the hill to a maximum height h = 3v²/4g. The moment of inertia of the object is (g= acceleration due to gravity)

Solution

Since it rolls without slipping, there is no friction. So, its initial mechanical energy at the horizontal surface equals its final mechanical energy at the top of the hill.

Since the object is rolling initially, and on horizontal ground, it initial energy is kinetic and made up of rotational and translational kinetic energy.

So, E = K + K'

E = 1/2mv² + 1/2Iω² where m = mass of object, v = speed of object, I = moment of inertia of object and ω = angular speed of object = v/r where v = speed of object and R = radius of object.

Also, the final mechanical energy of the object, E' is its potential energy at the top of the hill. So, E' = mgh.

Since E = E',

1/2mv² + 1/2Iω² = mgh

substituting the values of ω and h into the equation, we have

1/2mv² + 1/2Iω² = mgh

1/2mv² + 1/2I(v/R)²= mg(3v²/4g)

Expanding the brackets, we have

1/2mv² + 1/2Iv²/R²= 3mv²/4

Dividing through by v², we have

1/2m + I/2R²= 3m/4

Subtracting m/2 from both sides, we have

I/2R² = 3m/4 - m/2

Simplifying, we have

I/2R² = m/4

Multiplying through by 2R², we have

I = m/4 × 2R²

I = mR²/2

6 0

2 years ago

Fill in the blank. You don’t ____ me thass fine _ ___ -Man Tiffany

Jet001 [13]

Answer:

like, k bye

Hope this helps!

5 0

2 years ago

Read 2 more answers

Calculate the speed of a proton after it accelerates from rest through a potential difference of 350 V.

AVprozaik [17]

The speed of a proton after it accelerates from rest through a potential difference of 350 V is $25.86 \times 10^4 ~m/s$ .

Initial velocity of the proton $u = 0$

Given potential difference $\Delta V = 350V$

let's assume that the speed of the proton is $v$ ,

Since the proton is accelerating through a potential difference, proton's potential energy will change with time. The potential energy of a particle of charge $q$ when accelerated with a potential difference $\Delta V$ is,

$U = q \Delta V$

Due to Work-Energy Theorem and Conservation of Energy - If there is no non-conservative force acting on a particle then loss in Potential energy P.E must be equal to gain in Kinetic Energy K.E i.e

$\Delta K = \Delta V$