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Nesterboy [21]
1 year ago
14

Compared to the wavelength of incident light, diffuse reflection occurs when the size of surface irregularities is?

Physics
1 answer:
bagirrra123 [75]1 year ago
5 0

Compared to the wavelength of the incident light, diffuse reflection occurs when the size of surface irregularities is comparable to the size of the wavelength of the light.

The distance between the two crests or troughs of the light wave is known as the wavelength of light. Using the Greek letter lambda 'λ', it is identified. As a result, wavelength refers to the separation between one wave's crest or dip and the following wave.

In contrast to specular reflection, diffuse reflection involves the reflection of light, other waves, or particles off a surface so that rays incident on the surface are scattered at a variety of angles.

Refer to more about wavelength here

brainly.com/question/13533093

#SPJ4

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A positive charge of 18nC is evenly distributed along a straight rod of length 4.0 m that is bent into a circular arc with a rad
sergey [27]

Explanation:

Formula for angle subtended at the center of the circular arc is as follows.

           \theta = \frac{S}{r}

where,   S = length of the rod

              r = radius

Putting the given values into the above formula as follows.      

                \theta = \frac{S}{r}

                             = \frac{4}{2}

                             = 2 radians (\frac{180^{o}}{\pi})

                             = 114.64^{o}

Now, we will calculate the charge density as follows.

                 \lambda = \frac{Q}{L}

                            = \frac{18 \times 10^{-9} C}{4 m}

                            = 4.5 \times 10^{-9} C/m

Now, at the center of arc we will calculate the electric field as follows.

                 E = \frac{2k \lambda Sin (\frac{\theta}{2})}{r}

                     = \frac{2(9 \times 10^{9} Nm^{2}/C^{2})(4.5 \times 10^{-9}) Sin (\frac{114.64^{o}}{2})}{2 m}

                      = 34.08 N/C

Thus, we can conclude that the magnitude of the electric eld at the center of curvature of the arc is 34.08 N/C.

5 0
3 years ago
Lab number 14: sudden stops hurt newtons first law
kompoz [17]

Answer:

you want to know newtons first law here.

Explanation:

An object that is at rest stays at rest or stays in motion.

7 0
3 years ago
Why a boat cannot be made of solid steel.
Tju [1.3M]

Answer:

You want it to sink?

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5 0
3 years ago
Read 2 more answers
A 1.05 kg block slides with a speed of 0.865 m/s on a frictionless horizontal surface until it encounters a spring with a force
djyliett [7]

Answer:

a) U = 0 J    

k = 0.393 J

E = 0.393 J

b) U = 0.0229J

k = 0.370 J

E = 0.393 J

c) U = 0.0914 J

k = 0.302 J

E = 0.393 J

d) U = 0.206 J

k = 0.187 J

E = 0.393 J

e) U = 0.366 J

k = 0.027 J

E = 0.393 J

Explanation:

Hi there!

The equations of kinetic energy and elastic potential energy are as follows:

k = 1/2 · m · v²

U = 1/2 · ks · x²

Where:

m = mass of the block.

v = velocity.

ks = spring constant.

x = displacement of the string.

a) When the spring is not compressed, the spring potential energy will be zero:

U = 1/2 · ks · x²

U = 1/2 · 457 N/m · (0 cm)²

U = 0 J

The kinetic energy of the block will be:

k = 1/2 · m · v²

k = 1/2 · 1.05 kg · (0.865 m/s)²

k = 0.393 J

The mechanical energy will be:

E = k + U = 0.393 J + 0 J = 0.393 J

This energy will be conserved, i.e., it will remain constant because there is no work done by friction nor by any other dissipative force (like air resistance). This means that the kinetic energy will be converted only into spring potential energy (there is no thermal energy due to friction, for example).

b) The spring potential energy will be:

U = 1/2 · 457 N/m · (0.01 m)²

U = 0.0229 J

Since the mechanical energy has to remain constant, we can use the equation of mechanical energy to obtain the kinetic energy:

E = k + U

0.393 J = k + 0.0229 J

0.393 J - 0.0229 J = k

k = 0.370 J

c) The procedure is now the same. Let´s calculate the spring potential energy with x = 0.02 m.

U = 1/2 · 457 N/m · (0.02 m)²

U = 0.0914 J

Using the equation of mechanical energy:

E = k + U

0.393 J = k + 0.0914 J

k = 0.393 J - 0.0914 J = 0.302 J

d) U = 1/2 · 457 N/m · (0.03 m)²

U = 0.206 J

E = 0.393 J

k = E - U = 0.393 J - 0.206 J

k = 0.187 J

e) U = 1/2 · 457 N/m · (0.04 m)²

U = 0.366 J

E = 0.393 J

k = E - U = 0.393 J - 0.366 J = 0.027 J.

4 0
3 years ago
URGENT. Please help.
exis [7]

1). c ... 2). d ... 3). a ... 4). d ... 5). c ... 6). a

7). b-mass ... c-m/s ... d-Newton's 1st ... e-Newton's 2nd

6 0
3 years ago
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