5. What is the total length of the path of travel fron start of motion to the final place?

Light with energy equal to three times the work function of a given metal causes the metal to eject photoelectrons. What is the

Mrac [35]

The ratio of the maximum photoelectron kinetic energy to the work function will be 3:1.

<h3 /><h3>What is the photoelectric effect?</h3>

When a medium receives electromagnetic radiation, electrostatically charged particles are emitted from or inside it.

The emission of ions from a steel plate when light falls on it is a common definition of the effect. The substance could be a solid, liquid, or gas; and the released particles could be protons or electrons.

A particular metal emits photoelectrons when exposed to light with energy three times its work function:

$\rm KE=3 \phi$

The ratio of the maximum photoelectron kinetic energy to the work function will be;

$R=\frac{E}{\phi} \\\\ R=\frac{3 \phi}{\phi} \\\\ R= 3$

Hence, the ratio of the maximum photoelectron kinetic energy to the work function will be 3:1.

To learn more about the photoelectric effect refer to the link;

brainly.com/question/9260704

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5 0

1 year ago

A thin spherical glass shell in air is filled with an unknown liquid. A horizontal parallel light beam is incident on the shell

sammy [17]

I have the exact same question, any chance you figured it out since you posted this?

8 0

3 years ago

What would be the current of the same wire if it was moved up towards the magnet instead of through the magnetic field?

Aleks04 [339]

Answer:

0 v

Explanation:

3 0

3 years ago

The current through an inductor of inductance L is given by I(t) = Imax sin(ωt).

sammy [17]

Answer:

(a) $emf_L=-LI_{max}\omega cos(\omega t)$

(b) neither increasing or decreasing

(c) opposite to the flow of charge carriers

Explanation:

The current through an inductor of inductance L is given by:

$I(t)=I_{max}sin(\omega t)$ (1)

(a) The induced emf is given by the following formula

$emf_L=-L\frac{dI}{dt}$ (2)

You derivative the expression (1) in the expression (2):

$emf_L=-L\frac{d}{dt}(I_{max}sin(\omega t))\\\\emf_L=-LI_{max}\omega cos(\omega t)$

(b) At t=0 the current is zero

(c) At t = 0 the emf is:

$emf_L=-\omega LI_{max}$

w, L and Imax have positive values, then the emf is negative. Hence, the induced emf is opposite to the flow of the charge carriers.

(d) read the text carefully

6 0

3 years ago

Because of churning rock and ash, the largest volcanic eruptions are sometimes surrounded by

Sonja [21]

Calderas are what sometimes surround it

7 0

3 years ago