1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
aliya0001 [1]
1 year ago
11

_________ light is unique in that it vibrates mostly in one direction.

Physics
1 answer:
Misha Larkins [42]1 year ago
6 0

Polarized light light is unique in that it vibrates mostly in one direction.

Polarized mild waves are light waves in which the vibrations occur in a single plane. The process of reworking unpolarized light into polarized light is called polarization. There are a ramification of methods of polarizing mild.

Polarization is used for differentiating among transverse and longitudinal waves. Infrared spectroscopy makes use of polarization. it is used in seismology to take a look at earthquakes. In Chemistry, the chirality of organic compounds is tested the use of polarization strategies.

Refraction is the change in route of waves that takes place when waves tour from one medium to every other. . Light has a dual nature. as it has waves, sunlight passing via  rainstorm makes a rainbow. however, while mild moves a sun mobile, it can provide strength as a series of very small bursts. particles of count number have names together with the proton, electron and neutron.

Learn more about lights here:-brainly.com/question/19697218

#SPJ4

You might be interested in
Most chemical reactions can be divided into how many main groups?
tatyana61 [14]
There is synthesis
decomposition
double displacement
single displacement
combustion
metathesis
so i guess you could say 6
5 0
3 years ago
A high speed train is traveling at a speed of 44.7 m/s when the engineer sounds the 415 Hz warning horn. The speed of sound is 3
il63 [147K]

Answer:

Explanation:

Speed of the source of sound = v = 44.7 m/s

Speed of sound = V = 343 m/s

a) Apparent  frequency as the train approaches = f =  [V /(V -v) ] × f

= [343 / (343 - 44.7) ] × 415  = 477.18 Hz

Wave length =  λ = v / f = 343 / 477.18 = 0.719 m

b) Frequency heard as the train leaves = f ' =  [V / ( V + v) ] f

                                                                     = [343 / { 343 + 44.7 ) ] x 415

                                                                      = 367.2 Hz

Wavelength when leaving = v / f = 343 / 367.2 = 0.934 m

8 0
3 years ago
Phosphate never enters the<br> O atmosphere<br> O ground<br> O water<br> O ocean
gizmo_the_mogwai [7]
I think you can google this because I really don’t know the answer I’m so sorry
7 0
3 years ago
The following table lists the work functions of a few common metals, measured in electron volts. Metal Φ(eV) Cesium 1.9 Potassiu
Citrus2011 [14]

A. Lithium

The equation for the photoelectric effect is:

E=\phi + K

where

E=\frac{hc}{\lambda} is the energy of the incident light, with h being the Planck constant, c being the speed of light, and \lambda being the wavelength

\phi is the work function of the metal (the minimum energy needed to extract one photoelectron from the surface of the metal)

K is the maximum kinetic energy of the photoelectron

In this problem, we have

\lambda=190 nm=1.9\cdot 10^{-7}m, so the energy of the incident light is

E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{1.9\cdot 10^{-7} m}=1.05\cdot 10^{-18}J

Converting in electronvolts,

E=\frac{1.05\cdot 10^{-18}J}{1.6\cdot 10^{-19} J/eV}=6.5 eV

Since the electrons are emitted from the surface with a maximum kinetic energy of

K = 4.0 eV

The work function of this metal is

\phi = E-K=6.5 eV-4.0 eV=2.5 eV

So, the metal is Lithium.

B. cesium, potassium, sodium

The wavelength of green light is

\lambda=510 nm=5.1\cdot 10^{-7} m

So its energy is

E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{5.1\cdot 10^{-7} m}=3.9\cdot 10^{-19}J

Converting in electronvolts,

E=\frac{3.9\cdot 10^{-19}J}{1.6\cdot 10^{-19} J/eV}=2.4 eV

So, all the metals that have work function smaller than this value will be able to emit photoelectrons, so:

Cesium

Potassium

Sodium

C. 4.9 eV

In this case, we have

- Copper work function: \phi = 4.5 eV

- Maximum kinetic energy of the emitted electrons: K = 2.7 eV

So, the energy of the incident light is

E=\phi+K=4.5 eV+2.7 eV=7.2 eV

Then the copper is replaced with sodium, which has work function of

\phi = 2.3 eV

So, if the same light shine on sodium, then the maximum kinetic energy of the emitted electrons will be

K=E-\phi = 7.2 eV-2.3 eV=4.9 eV

7 0
3 years ago
During the fission reaction shown, how did the target nucleus change ?
Zarrin [17]

Answer:

A. The target nucleus split into two nuclei, each with fewer nucleons than the original.

Explanation:

5 0
3 years ago
Other questions:
  • three eggs, each with a mass of ten grams are cooked to make scrambled eggs, what is the total mass of the scrambled eggs​
    5·1 answer
  • A 1500-kg car accelerates from rest to 25 m/s in 7.0 s. what is the average power delivered by the engine? (1 hp = 746 w)
    15·2 answers
  • The diagram below shows a subduction zone where oceanic crust is sinking into the mantle underneath continental crust as two
    13·1 answer
  • Great amounts of heat are found at hotspots. What is the source of this heat?
    14·1 answer
  • The _______ is/are a thin sheet of skin at the end of the outer ear canal that vibrates in response to sound.
    8·1 answer
  • A ball hits a wall. What is true about the magnitude of the force experienced by the ball compared with the force experienced by
    6·1 answer
  • PLEASE HELP ASAP!! DUE IN AN HOUR :’))!!
    14·1 answer
  • What is a electric curcit
    5·2 answers
  • The type of torque wrench designed for tightening clamping bands on underground pipe is the ___________________
    15·2 answers
  • What is the total distance, side to side, that the top of the building moves during such an oscillation
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!