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3 years ago

What is the best description of the destructive interference of light?

Physics

1 answer:

timama [110]3 years ago

3 0

Answer:

Sorry if I am late but its A. The crest of one wave overlaps with the trough of another.

Explanation:

Just trust me lol and its on k12

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Help i forgot to complete this worksheet

Naily [24]

just search up the answer/ definition to all of them, rephrase into own words, then do the same for examples.

6 0

3 years ago

When light of wavelength 240 nm falls on a cobalt surface, electrons having a maximum kinetic energy of 0.17 eV are emitted. Fin

dusya [7]

Answer:

(a) 5.04 eV (B) 248.14 nm (c) $1.21\times 10^{15}Hz$

Explanation:

We have given Wavelength of the light \lambda = 240 nm

According to plank's rule ,energy of light

$E = h\nu = \frac{hc}{}\lambda$

$E = h\nu = \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{ 240\times 10^{-9} m\times 1.6\times 10^{-19}J/eV}= 5.21 eV$

Maximum KE of emitted electron i= 0.17 eV

Part( A) Using Einstien's equation

$E = KE_{max}+\Phi _{0}$ , here $\Phi _0$ is work function.

$\Phi _{0}=E - KE_{max}$ = 5.21 eV-0.17 eV = 5.04 eV

Part( B) We have to find cutoff wavelength

$\Phi _{0} = \frac{hc}{\lambda_{cuttoff}}$

$\lambda_{cuttoff}= \frac{hc}{\Phi _{0} }$

$\lambda_{cuttoff}= \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{5.04 eV\times 1.6\times 10^{-19}J/eV }=248.14 nm$

Part (C) In this part we have to find the cutoff frequency

$\nu = \frac{c}{\lambda_{cuttoff}}= \frac{3\times 10^{8}m/s}{248.14 \times 10^{-19} m }= 1.21\times 10^{15} Hz$

5 0

3 years ago

3. Read this sentence from the text: "Most of the solar radiation is absorbed by the atmosphere

Semmy [17]

I think the answer is A

6 0

3 years ago

Which will produce a magnetic field?

Yanka [14]

Answer:

a magnet

3 0

3 years ago

Two balloons are charged with an identical quantity and type of charge: -6.25 x 10-6 C. They are held apart at a separation dist

steposvetlana [31]

Answer:

The electric force between them is 878.9 N

Explanation:

Given:

Identical charge $q = -6.25 \times 10^{-6}$ C

Separation between two charges $r = 0.02$ m

For finding the electrical force,

According to the coulomb's law

$F = \frac{k q^{2} }{r^{2} }$

Here, force between two balloons are repulsive because both charges are same.

Where $k = 9 \times 10^{9}$

$F = \frac{9 \times 10^{9} \times (-6.25 \times 10^{-6} )^{2} }{4 \times 10^{-4} }$

$F = 878.9$ N

Therefore, the electric force between them is 878.9 N

8 0

3 years ago