All categories

3 years ago

Refraction occurs when a wave

Physics

2 answers:

lesya692 [45]3 years ago

6 0

Answer:

Explanation:

Did on a quiz and got it right

sukhopar [10]3 years ago

4 0

The answer is A because refraction means radio waves or the way the light hits the waves which causes it to be deflected and change direction which is exactly what a wave does

You might be interested in

I don't understand this question at all, can I please get some help?

beks73 [17]

V^2/R=180W
v=root 180R

4 0

3 years ago

A double-slit experiment is set up using red light (λ = 706 nm). A first order bright fringe is seen at a given location on a sc

Elanso [62]

Answer:

λ = 470.66 nm

Explanation:

for bright fringe $y_m = \frac{m\lambda D}{d}$

D= distance between slit and screen

d= distance between the slits

for first order bright fringe m = 1,

$y_1 = \frac{1\lambda D}{d}$

$y_1 = {706*D}{d}$

for dark fringe,we have

$y_m = {(m + 1/2)\lambda D}{d}$

Now to get the dark fringes at the same location we should have;

(706)D/d = (m + 1/2)λD/d

put m = 1

(1 + 1/2)λ = (706)

λ = 470.66 nm

6 0

3 years ago

How do high-energy electrons from glycolysis and the krebs cycle contribute to the formation of atp from adp in the electron tra

ivanzaharov [21]

Answer

Together with glycolysis, The Krebs cycle, and the electron transport chain release about 36 molecules of ATP per molecule of glucose.The Krebs cycle uses the two molecules of pyruvic acid formed in glycolysis and yields high-energy molecules of NADH and flavin adenine dinucleotide (FADH2), as well as some ATP. The electron transport chain forms a proton gradient across the inner mitochondrial membrane, which drives the synthesis of ATP

5 0

3 years ago

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:

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

Light enters water from air at an angle of 25° with the normal, Θ1. If water has an index of refraction of 1.33, determine Θ2.

vladimir1956 [14]

By Snell's law:

η = sini / sinr. i = 25, η = 1.33

1.33 = sin25° / sinr

sinr = sin25° / 1.33 = 0.4226/1.33 = 0.3177 Use a calculator.

r = sin⁻¹(0.3177)

r ≈ 18.52°

Option A.

God's grace.

6 0

3 years ago