If a body starts moving from rest its<br>initial velocity is<br>ų <br>V<br>zero<br>m/s

Isaac Newton used a ____ to break white light into its component colors.

ivann1987 [24]

<h2>Answer: Prism</h2>

In the eighteenth century Isaac Newton found out that <u>when a beam of light from the Sun, passes trhough a prism is decomposed in many different colors</u>. He named this phenomenom as dispersion of light.

This phenomenom occurs when a beam of white light (which is compound of many wavelengths or "colors") is refracted (the different rays of light are diverted depending on their wavelengths) in some medium, leaving their constituent colors separated.

Therefore:

<h2>Isaac Newton used a <u>prism</u> to break white light into its component colors.</h2>

8 0

3 years ago

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An object has a coefficient of static friction of 0.3 and a normal force of 30 N. Find the force of static friction.

gladu [14]

Answer:

9N

Explanation:

static friction=normal force x coefficient of static friction

so static friction =30N x 0.3= 9N

7 0

3 years ago

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g How many diffraction maxima are contained in a region of the Fraunhofer single-slit pattern, subtending an angle of 2.12°, for

lbvjy [14]

Answer:

2

Explanation:

We know that in the Fraunhofer single-slit pattern,

maxima is given by

$a\text{sin}\theta=\frac{2N+1}{2}\lambda$

Given values

θ=2.12°

slit width a= 0.110 mm.

wavelength λ= 582 nm

Now plugging values to calculate N we get

$0.110\times10^{-3}\text{sin}2.12=(\frac{2N+1}{2})582\times10^{-9}$

Solving the above equation we get

we N= 2.313≅ 2

4 0

3 years ago

A monochromatic light beam is incident on a barium target that has a work function of 2.50 \mathrm{eV} . If a potential differen

leva [86]

The wavelength of the light beam required to turn back all the ejected electrons is 497 nm which is option (b).

Work function is a material property defined as the minimum amount of energy required to infinitely remove electrons from the surface of a particular solid.
The potential difference required to support all emitted electrons is called the stopping potential which is given by $v_0=\frac{K.E_m_a_x}{e}$ .....(1)
where $v_0$ is the stopping potential and e is the charge of the electron given by $1.6\times10^-^1^9$ .

It is given that work function (Ф) of monochromatic light is 2.50 eV.

Einstein photoelectric equation is given by:

$K.E_m_a_x=E-\phi$ ....(2)

where K.E(max) is the maximum kinetic energy.

Substituting (1) into (2) , we get

$ev_0=E-\phi\\1.6\times10^{-19} \times1=E-2.50\\E=1.6\times10^{-19}+2.50\\E=2.50eV$

As we know that $E=\frac{hc}{\lambda}$ ....(3)

where Speed of light, $c = 3\times10^8 m/s$ and Planck's constant , $h = 6.63\times 10^-^1^9Js = 4.14\times 10^-^1^5 eVs$

From equation (3) , we get

$\lambda=\frac{hc}{E} \\\\\lambda=\frac{ 4.14\times 10^-^1^5 \times 3 \times10^8}{2.50} \\\\\lambda=\frac{1240\times10^-^9}{2.50} \\\\\lambda=496.8\times10^-^9\\\\\lambda=497nm$