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Anna11 [10]
3 years ago
6

What are the like terms in the expression? 4m-9+3n+2

Physics
2 answers:
mash [69]3 years ago
8 0
The like terms in this expression are, -9 and 2, but that is it
wel3 years ago
5 0
-9 and +2
Hope I helped :)

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Unpolarized light is passed through an optical filter that is oriented in the vertical direction.
VashaNatasha [74]

In order to solve this problem it is necessary to apply the concepts related to intensity and specifically described in Malus's law.

Malus's law warns that

I = I_0 cos^2\theta

Where,

\theta= Angle between the analyzer axis and the polarization axis

I_0 =Intensity of the light before passing through the polarizer

The intensity of the beam from the first polarizer is equal to the half of the initial intensity

I = \frac{I_0}{2}

Replacing with our the numerical values we get

I = \frac{46}{2}

I = 23W/m^2

Therefore the  intensity of the light that emerges from the filter is 23W/m^2

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kotegsom [21]

We can answer this one very quickly.  From the <em>Law of Conservation of Energy</em>, we know that "Energy can't be created or destroyed.".

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"One form of energy can be <em>transformed into</em> another type of energy. "  <em>(B)</em>

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Do you think an observer on another planet in the solar system might see eclipses
artcher [175]

Answer:

Yes

Explanation:

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Talking about the Earth, the geometry is such that the Moon and the Sun are of same apparent size as seen from the Earth. Thus Lunar and Solar eclipse can be seen from the Earth. If we were to go on any other planet the same phenomenon can be seen provided the apparent size of moon and the Sun from that planet is same.

We have seen and recorded many such eclipses on Jupiter. These are from the perspective of Earth. When the moons of Jupiter comes exactly between the Sun and Jupiter the shadow of moon will fall on Jupiter. The places where the shadow falls, one will see a solar eclipse.

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Find the kinetic energy of an electron whose de broglie wavelength is 34.0 nm.
dezoksy [38]
The De Broglie wavelength of the electron is
\lambda=34.0 nm=34 \cdot 10^{-9} m
And we can use De Broglie's relationship to find its momentum:
p= \frac{h}{\lambda}= \frac{6.6 \cdot 10^{-34} Js}{34 \cdot 10^{-9} m}=1.94 \cdot 10^{-26} kg m/s

Given p=mv, with m being the electron mass and v its velocity, we can find the electron's velocity:
v= \frac{p}{m}= \frac{1.94 \cdot 10^{-26} kgm/s}{9.1 \cdot 10^{-31} kg}=  2.13 \cdot 10^4 m/s

This velocity is quite small compared to the speed of light, so the electron is non-relativistic and we can find its kinetic energy by using the non-relativistic formula:
K= \frac{1}{2}mv^2= \frac{1}{2}(9.1 \cdot 10^{-31} kg)(2.13 \cdot 10^4 m/s)^2=2.06 \cdot 10^{-22} J
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3 years ago
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