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EleoNora [17]
4 years ago
9

What does periodic mean

Physics
1 answer:
Anvisha [2.4K]4 years ago
4 0

Answer:

Something that relates to the periodic table

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Which type of wave would not be classified as a transverse wave?
Jlenok [28]
Longitud wave something like that.
6 0
4 years ago
You have an unpolarized light source and you wish to send a beam of this light with intensity Io through a number of sheets of p
tekilochka [14]

Answer:

Explanation:

a ) The angle  between the polarization axis of two adjecent sheet

= 90 / 3 = 30 degree.

The formula for intensity of polarised light from unpolarised light ( first transmission

I₁ = I₀ /2

I₀ is intensity of unpolarised light and I₁ is intensity of light after first time polarization .

The relation of I₁ and I₂ is as follows

I₂ = I₁ cos²30

= I₀/2 x3/4

=3 I₀/8

Relation between I₃ and I₂ is as follows

I₃ = I₂ cos²30

= 3I₀ / 8 x 3/4

= 9 I₀ / 32

= 0 .28 I₀

In case of stack of 4 plates

angle between two plates  = 90/4  = 22.5 degree

I₁ = I₀ /2

I₂ = I₁ cos²22.5

=  I₀ /2 x .85

I₃ = I₂ cos²22.5

= I₀ /2 x .85 x .85

= .36 I₀

7 0
3 years ago
Katrina’s recipe for orange muffins calls for 400 grams of flour. How many kilograms of flour are in the recipe?
aliina [53]

Answer:

0.4 Kilograms of flour

Explanation:

If you convert 400 grams into kilograms (400 ÷ 1,000), you would get 0.4 kilograms.

3 0
3 years ago
Which layer of the sun is responsible for producing the light shown in the picture above?
Leokris [45]
Chromosphere is the layer that is responsible for producing the sunlight

8 0
3 years ago
Read 2 more answers
Calculate the deflection at point C of a beanm subjected to uniformly distributed load w 275 N/mm on span AB and point load P-10
lakkis [162]

Answer:

Deflection at point C = 0.151 m

Explanation:

Central deflection due to uniformly distributed load w for simply supported beam

         \delta =\frac{5wl^4}{384EI}

Central deflection due to point load P for simply supported beam

         \delta =\frac{Pl^3}{48EI}

Here deflection

          \delta =\frac{5wl^4}{384EI}+\frac{Pl^3}{48EI}=\frac{5\times 275\times 10^3\times 5^4}{384\times 1.5\times 10^7}+\frac{10\times 10^3\times 5^3}{48\times 1.5\times 10^7}\\\\\delta =0.149 + 1.74\times 10^{-3}=0.151m

Deflection at point C = 0.151 m

7 0
4 years ago
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