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3 years ago

HELP ME PLEASE

Physics

2 answers:

dimulka [17.4K]3 years ago

8 0

Answer:

Wax paper and frosted glass are translucent materials.

Explanation:

You can usually see something behind a translucent object, but the details are blurred.

Vlada [557]3 years ago

5 0

Answer: A part of the incident light may get reflected or scattered, as it passes through the interior of the material. Any object, seen through a translucent material, appears fuzzy or blurred. Some examples are oily paper, tissue, some plastics, etc. [Image will be Uploaded Soon]

Explanation:

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He distance between two consecutive crests is 2.5 meters. Which characteristic of the wave does this distance represent? amplitu

kupik [55]

Frequency is measured in units of reciprocal time.
Period is measured in units of time.
Phase is a number without units that represents a fraction of a wave.

None of these is measured in meters, so none of them can be the answer.
It must be either amplitude or wavelength.

Amplitude is a quantity that's measured at one or two points in the same wave.
The question is talking about points on consecutive waves.

<em>Wavelength is</em> the only choice left. That must be it.

3 0

2 years ago

Read 2 more answers

The average intensity of light emerging from a polarizing sheet is 0.708 W/m2, and that of the horizontally polarized light inci

Pachacha [2.7K]

Answer:

Angle θ = 30.82°

Explanation:

From Malus’s law, since the intensity of a wave is proportional to its amplitude squared, the intensity I of the transmitted wave is related to the incident wave by; I = I_o cos²θ

where;

I_o is the intensity of the polarized wave before passing through the filter.

In this question,

I is 0.708 W/m²

While I_o is 0.960 W/m²

Thus, plugging in these values into the equation, we have;

0.708 W/m² = 0.960 W/m² •cos²θ

Thus, cos²θ = 0.708 W/m²/0.960 W/m²

cos²θ = 0.7375

Cos θ = √0.7375

Cos θ = 0.8588

θ = Cos^(-1)0.8588

θ = 30.82°

4 0

3 years ago

Help Please

hichkok12 [17]

Gravitational potential energy i think

8 0

3 years ago

a solid metal sphere of radius 3.00m carries a total charge of -5.50. what is the magnitude of the electric field at a distance

aivan3 [116]

Answer:

(a) Electric field at 0.250 m is zero.

(b) Electric field at 2.90 m is zero.

(c) Electric field at 3.10 m is - 5.15 x 10³ V/m.

(d) Electric field at 8.00 m is - 0.77 x 10³ V/m.

Explanation:

Let Q and R are the charge and radius of the solid metal sphere. The solid metal sphere behave as conductor, so total charge Q is on the surface of the sphere.

Electric field inside and outside the metal sphere is :

E = 0 for r ≤ R ( inside )

= $\frac{KQ}{r^{2} }$ for r > R ( outside )

Here K is electric constant and r is the distance from the center of the metal sphere.

(a) Electric field at 0.250 m is zero as r < R i.e. 0.250 m < 3 m from the above equation.

(b) Electric field at 2.90 m is zero as r < R i.e. 2.90 m < 3 m from the above equation.

E = $\frac{KQ}{r^{2} }$

Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 3.10 m for r in the above equation.

E = - $\frac{9\times10^{9}\times5.50\times10^{-6} }{3.10^{2} }$

E = - 5.15 x 10³ V/m

(d) Electric field at 8.00 m is given by the relation as r > R :

E = $\frac{KQ}{r^{2} }$

Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 8.00 m for r in the above equation.

E = - $\frac{9\times10^{9}\times5.50\times10^{-6} }{8^{2} }$

E = - 0.77 x 10³ V/m

8 0

3 years ago

Does every light source emit only one type of light?

blsea [12.9K]

Yes it is possible. Spectrum of emitted light depends upon the chemical composition of the source. and the way of its excitation. a clear example to us is that of sun.

3 0

2 years ago