Answer:
θ = 32.4º
Explanation:
For this exercise let's use Malus's law
I = Io cos² θ
in this case it indicates that the incident intensity is 370 W/m², when the first polarization passes, only the radiation with the same polarization of the polarizer emerges, that is, vertical
I₀ = 370/2 = 185 W / m²
this is the radiation that affects the second polarizer, let's apply the expression of Maluz
θ = cos⁻¹ ([tex]\sqrt{\frac{I}{I_o} }[/tex])
θ = cos⁻¹ ([tex]\sqrt{132/185}[/tex])
θ = cos⁻¹ (0.844697)
θ = 32.4º
Scientists are constantly exploring the universe, looking for new planets that support life similar to the life on
Earth. A new planet that supports life would have all of the following characteristics except -
A. a gaseous atmosphere.
B. an orbiting moon.
C. liquid water.
D. protection from radiation.
A new planet that supports life would have all the following characteristics except an orbiting moon. Hence, option B is correct.
What is a Planet?An enormous, spherical celestial object known as a planet is neither a star nor its remains. The nebular hypothesis, which states how an interstellar cloud falls out of a nebula to produce a young protostar encircled by a protoplanetary disk, is now the best explanation for planet formation.
By gradually accumulating material under the influence of gravity, or accretion, planets develop in this disk.
The rocky planets Mercury, Venus, Earth, and Mars, as well as the giant planets Jupiter, Saturn, Uranus, and Neptune, make up the Solar System's minimum number of eight planets. These planets all revolve around axes that are inclined relative to their respective polar axes.
To know more about Planet:
https://brainly.com/question/14581221
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15 points!
a. Calculate the electric potential energy stored in a 1.4 x 10-7 F capacitor
that stores 3.40 x 10-6 C of charge at 24.0 V.
Answer:
[tex]4.12\times 10^{-5}\ J[/tex].
Explanation:
Given that,
Capacitance, [tex]C=1.4\times 10^{-7}\ F[/tex]
Charge stored in the capacitor, [tex]Q=3.4\times 10^{-6}\ C[/tex]
We need to find the electric potential energy stored in the capacitor. The formula for the electric potential energy stored in the capacitor is given by :
[tex]E=\dfrac{Q^2}{2C}[/tex]
Put all the values,
[tex]E=\dfrac{(3.4\times 10^{-6})^2}{2\times 1.4\times 10^{-7}}\\\\=4.12\times 10^{-5}\ J[/tex]
So, the required electric potential eenergy is equal to [tex]4.12\times 10^{-5}\ J[/tex].