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3 years ago

I am unable to understand from question no 3 and I have assignment due :( in 2 days need to do till 9

Physics

1 answer:

ludmilkaskok [199]3 years ago

8 0

Answer:

Oh well have your parents do something bout that. They are there to help you. Or guardian they all shall help you that is what they are there for. They are they to take care of you, answer you questions, and last but not least, they are there to also raise you. Ask you mom, dad, guardian, and maybe even you teacher to post pone the date. Teachers are to help you. Your guardian is to watch you make sure your safe while your parents are away (bascially like raising you). And parents to raise you, make sure you safe, and many other things like to make sure you are fed.

Explanation:

I hope they all have some type of help for you.

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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

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

Learn about more einstein photoelectric equation here:

brainly.com/question/11683155

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8 0

1 year ago

An object is 39 cm away from a concave mirrors surface along the principles axis. If the mirrors focal length is 9.50 cm, how fa

Tatiana [17]

Answer:

12.6 cm

Explanation:

We can use the mirror equation to find the distance of the image from the mirror:

$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$

where here we have

f = 9.50 cm is the focal length

p = 39 cm is the distance of the object from the mirror

Solving the equation for q, we find:

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{9.50 cm}-\frac{1}{39 cm}=0.080 cm^{-1}\\q = \frac{1}{0.080 cm}=12.6 cm$

5 0

3 years ago

Please help

adelina 88 [10]

Answer:

I'm pretty sure that the answer is A

5 0

3 years ago

Explain 2 reasons why reactions take place faster at high temperature

Vlada [557]

I have one reason the reaction take place faster because the molecules are going at a faster pace because the temperature is rising

8 0

3 years ago

Read 2 more answers

Consider a cyclotron in which a beam of particles of positive charge q and mass m is moving along a circular path restricted by

Ulleksa [173]

A) $v=\sqrt{\frac{2qV}{m}}$

B) $r=\frac{mv}{qB}$

C) $T=\frac{2\pi m}{qB}$

D) $\omega=\frac{qB}{m}$

E) $r=\frac{\sqrt{2mK}}{qB}$

Explanation:

When the particle is accelerated by a potential difference V, the change (decrease) in electric potential energy of the particle is given by:

$\Delta U = qV$

where

q is the charge of the particle (positive)

On the other hand, the change (increase) in the kinetic energy of the particle is (assuming it starts from rest):

$\Delta K=\frac{1}{2}mv^2$

where

m is the mass of the particle

v is its final speed

According to the law of conservation of energy, the change (decrease) in electric potential energy is equal to the increase in kinetic energy, so:

$qV=\frac{1}{2}mv^2$

And solving for v, we find the speed v at which the particle enters the cyclotron:

$v=\sqrt{\frac{2qV}{m}}$

When the particle enters the region of magnetic field in the cyclotron, the magnetic force acting on the particle (acting perpendicular to the motion of the particle) is

$F=qvB$

where B is the strength of the magnetic field.

This force acts as centripetal force, so we can write:

$F=m\frac{v^2}{r}$

where r is the radius of the orbit.

Since the two forces are equal, we can equate them:

$qvB=m\frac{v^2}{r}$

And solving for r, we find the radius of the orbit:

$r=\frac{mv}{qB}$ (1)

The period of revolution of a particle in circular motion is the time taken by the particle to complete one revolution.

It can be calculated as the ratio between the length of the circumference ( $2\pi r$ ) and the velocity of the particle (v):

$T=\frac{2\pi r}{v}$ (2)

From eq.(1), we can rewrite the velocity of the particle as

$v=\frac{qBr}{m}$

Substituting into(2), we can rewrite the period of revolution of the particle as:

$T=\frac{2\pi r}{(\frac{qBr}{m})}=\frac{2\pi m}{qB}$

And we see that this period is indepedent on the velocity.

The angular frequency of a particle in circular motion is related to the period by the formula

$\omega=\frac{2\pi}{T}$ (3)

where T is the period.

The period has been found in part C:

$T=\frac{2\pi m}{qB}$

Therefore, substituting into (3), we find an expression for the angular frequency of motion:

$\omega=\frac{2\pi}{(\frac{2\pi m}{qB})}=\frac{qB}{m}$

And we see that also the angular frequency does not depend on the velocity.

For this part, we use again the relationship found in part B:

$v=\frac{qBr}{m}$

which can be rewritten as

$r=\frac{mv}{qB}$ (4)

The kinetic energy of the particle is written as

$K=\frac{1}{2}mv^2$

So, from this we can find another expression for the velocity:

$v=\sqrt{\frac{2K}{m}}$

And substitutin into (4), we find:

$r=\frac{\sqrt{2mK}}{qB}$

So, this is the radius of the cyclotron that we must have in order to accelerate the particles at a kinetic energy of K.

Note that for a cyclotron, the acceleration of the particles is achevied in the gap between the dees, where an electric field is applied (in fact, the magnetic field does zero work on the particle, so it does not provide acceleration).

6 0

3 years ago