Calculate the work done in lifting a 500-N

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kogti [31]

Answer:

C. nuclear to electrical

6 0

3 years ago

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Electromagnetic radiation of 8.12×10¹⁸ Hz frequency is applied on a metal surface and caused electron emission. Determine the wo

klemol [59]

Answer:

The work function ϕ of the metal = 53.4196 x 10⁻¹⁶ J

Explanation:

When light is incident on a photoelectric material like metal, photoelectrons are emitted from the surface of the metal. This process is called photoelectric effect.

The relationship between the maximum kinetic energy ( $E_{k}$ ) of the photoelectrons to the frequency of the absorbed photons (f) and the threshold frequency (f₀) of the photoemissive metal surface is:

$E_{k}$ = h(f − f₀)

$E_{k}$ = hf - hf₀

E is the energy of the absorbed photons: E = hf

ϕ is the work function of the surface: ϕ = hf₀

$E_{k}$ = E - ϕ

Frequency f = 8.12×10¹⁸ Hz

Maximum kinetic energy $E_{k}$ = 4.16×10⁻¹⁷ J

Speed of light c = 3 x 10⁸ m/s

Planck's constant h = 6.63 × 10⁻³⁴ Js

E = hf = 6.63 × 10⁻³⁴ x 8.12×10¹⁸

E = 53.8356 x 10⁻¹⁶ J

from $E_{k}$ = E - ϕ ;

ϕ = E - $E_{k}$

ϕ = 53.8356 x 10⁻¹⁶ - 4.16×10⁻¹⁷

ϕ = 53.4196 x 10⁻¹⁶ J

The work function of the metal ϕ = 53.4196 x 10⁻¹⁶ J

4 0

3 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

4 years ago

Light of wavelength 633 nm falls on a single slit 0.5 mm wide and forms a diffraction pattern on a screen 1.0 m away from the sl

andrey2020 [161]

Answer:

The position of the first dark spot on the positive side of the central maximum is 1.26 mm.

Explanation:

Given that,

Wavelength of light is 633 nm.

Slit width, d = 0.5 mm

The diffraction pattern forms on a screen 1 m away from the slit. We need to find the position of the first dark spot on the positive side of the central maximum.

For destructive interference :

$\dfrac{dY}{D}=n\lambda$

Y is the distance of the minima from central maximum

Here, n = 1

$Y=\dfrac{n\lambda D}{d}\\\\Y=\dfrac{1\times 633\times 10^{-9}\times 1}{0.5\times 10^{-3}}\\\\Y=0.00126\ m\\\\Y=1.26\times 10^{-3}\ m\\\\Y=1.26\ mm$

So, the position of the first dark spot on the positive side of the central maximum is 1.26 mm.

4 0

3 years ago

What is a diffraction grating? What is a diffraction grating? A device used to grate cheese and other materials A musical instru

hammer [34]

Answer:

An opaque object with many closely spaced slits

Explanation:

Diffraction grating is an opaque object which will have many close slits on it

Each slit will allow the light wave pass through it which incident on it.

So here each slit will behave like a secondary source which will transmit the light and the superposition of light is then observed on the screen

These large number of slits on the object is combined known as diffraction grating.

By the superposition of waves due to grating we will have pattern of maximum and minimum intensity on the screen and this intensity if highest at the mid point of screen and then decreases as we move away

5 0

3 years ago