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Hitman42 [59]
3 years ago
6

In which state of matter has the LEAST kinetic energy?

Physics
2 answers:
Debora [2.8K]3 years ago
5 0

Answer:

the answer is solid bro

serious [3.7K]3 years ago
3 0
A solid state because the particles are tightly packed and particles are not able to make freely.
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How are the forces observed related to the motion of the earth around the sun
Inga [223]

This question sounds like it came after some activity where
some forces were observed.  Since we were not there, and
we don't know what the activity was, we don't know what forces
were observed, and we have no clue to how they might be related
to the motion of the Earth around the sun.

7 0
3 years ago
Explain why asleel meedle does not csink on water​
Juli2301 [7.4K]

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upthrust

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5 0
3 years ago
Controlling the amount of current in a circuit by opposing the flow of charge
Vera_Pavlovna [14]

Answer:

Resistor

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8 0
3 years ago
Read 2 more answers
A grating whose slits are 3.2x10^-4 cm apart produces a third-order fringe at a 25.°0 angle. What is the wavelength of light tha
Ber [7]

Answer:

The light used has a wavelenght of 4.51×10^-7 m.

Explanation:

let:

n be the order fringe

Ф be the angle that the light makes

d is the slit spacing of the grating

λ be the wavelength of the light

then, by Bragg's law:

n×λ = d×sin(Ф)

λ = d×sin(Ф)/n

λ = (3.2×10^-4 cm)×sin(25.0°)/3

  = 4.51×10^-5 cm

  ≈ 4.51×10^-7 m

Therefore, the light used has a wavelenght of 4.51×10^-7 m.

7 0
3 years ago
Unpolarized light with intensity I0I0I_0 is incident on an ideal polarizing filter. The emerging light strikes a second ideal po
zvonat [6]

Answer:

0.293I_0

Explanation:

When the unpolarized light passes through the first polarizer, only the component of the light parallel to the axis of the polarizer passes through.

Therefore, after the first polarizer, the intensity of light passing through it is halved, so the intensity after the first polarizer is:

I_1=\frac{I_0}{2}

Then, the light passes through the second polarizer. In this case, the intensity of the light passing through the 2nd polarizer is given by Malus' law:

I_2=I_1 cos^2 \theta

where

\theta is the angle between the axes of the two polarizer

Here we have

\theta=40^{\circ}

So the intensity after the 2nd polarizer is

I_2=I_1 (cos 40^{\circ})^2=0.587I_1

And substituting the expression for I1, we find:

I_2=0.587 (\frac{I_0}{2})=0.293I_0

5 0
3 years ago
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