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3 years ago

How are the Northern Lights are formed.

Physics

1 answer:

Irina-Kira [14]3 years ago

5 0

Answer:

<h3>Bottom line: When charged particles from the sun strike atoms in Earth's atmosphere, they cause electrons in the atoms to move to a higher-energy state. When the electrons drop back to a lower energy state, they release a photon: light. This process creates the beautiful aurora, or northern lights.</h3>

Explanation:

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The average intensity of light emerging from a polarizing sheet is 0.775 W/m2, and that of the horizontally polarized light inci

rewona [7]

Answer:

the transmission axis of polarizing sheet makes an angle of $\Theta =26.74^{\circ}$ with the horizontal

Explanation:

We have given that intensity of light incident on the sheet $I_0=0.970W/m^2$

Average intensity of light emerging from a polarizing sheet $I=0.775W/m^2$

We have to find the angle between transmission axis with the horizontal

Intensity of light polarizing from sheet is equal to $I=I_0cos^2\Theta$

So $0.775=0.970cos^2\Theta$

$cos^2\Theta =0.798$

$cos\Theta =0.893$

$\Theta =26.74^{\circ}$

So the transmission axis of polarizing sheet makes an angle of $\Theta =26.74^{\circ}$ with the horizontal

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3 years ago

If sugar contains 41.86% carbon and 6.98% hydrogen what percentage of sugar is oxygen?

MrMuchimi

If the only thing remaining is Oxygen then it is 51.16%.

4 0

4 years ago

Estimate the average power output of the sun, given that about 1350 w/m2 reaches the upper atmosphere of the earth.

fenix001 [56]

the <u>average powe</u><u>r</u> output of the sun is <u>3.8 × </u> $10^{26}$ <u> W</u> given that about 1350 w/m2 reaches the upper atmosphere of the earth.

As we know intensity falling over a surface is equal to Energy falling per unit area per unit volume.

I = $\frac{Energy}{area*time}$

Since Power is equal to energy/time

I = power/area

Power = Intensity × area

Now, Distance from Sun to Earth , r = 149.6 × $10^{9}$ m

So, Surface area of the sphere of radius is:

A = 4 π $r^{2}$

= 4 × 3.14 × (149.6 × $10^{9}$ )²

= 2.81 × $10^{23}$ m²

Thus Average Power output of the sun will be:

P = 1350 × 2.81 × $10^{23}$

= 3.7935 × $10^{26}$

P = 3.8 × $10^{26}$ W

So the average power output of the sun is 3.8 × $10^{26}$ W.

If you need to learn more about about the average power of a resistor, click here.

brainly.com/question/12972958

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7 0

2 years ago

Read 2 more answers

A charge of -8.00 nC is spread uniformly over the surface of one face of a nonconducting disk of radius 1.05 cm.

gavmur [86]

Answer:

(a) $E = -1.02 \times 10^5~N/C$

(b) $E = -9.7 \times 10^4~N/C$

Explanation:

(a) The electric field for a point charge is given by the following formula:

$\vec{E} = \frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}\^r$

Since this formula is valid for point charges, we have to choose an infinitesimal area, da, from the disk. Then we will calculate the E-field (dE) created by this small area using the above formula, then we will integrate over the entire disk to find the E-field created by the disk.

$dE = \frac{1}{4\pi\epsilon_0}\frac{dQ}{(\sqrt{z^2 + r^2})^2}$

Here, z = 0.025 m. And r is the distance of the infinitesimal area from the axis. dQ is the charge of the small area, and should be written in terms of the given variables.

In cylindrical coordinates, da = r dr dθ. So,

$\frac{Q}{\pi R^2} = \frac{dQ}{da}\\\frac{Q}{\pi R^2} = \frac{dQ}{rdrd\theta}\\dQ = \frac{Qrdrd\theta}{\pi R^2}$

Hence, dE is now:

$dE = \frac{1}{4\pi\epsilon_0}\frac{Q}{\pi R^2}\frac{rdrd\theta}{z^2 + r^2}$

The surface integral over the disk can now be taken, but there is one more thing to be considered. This dE is a vector quantity, and it needs to be separated its components.

It has two components, one in the vertical direction and another in the horizontal direction. By symmetry, the horizontal components cancel out each other in the end (since it is a disk, each horizontal vector has an equal but opposite counterpart), so only the vertical component should be considered.

Let us denote the angle between dE and the horizontal axis as α. This angle can be found by the geometry of the triangle formed by dE, vertical axis of the disk, and horizontal plane. So,

$\sin(\alpha) = \frac{z}{\sqrt{z^2 + r^2}}$

Therefore, vertical component of dE now becomes

$dE_z = \frac{1}{4\pi\epsilon_0}\frac{Q}{\pi R^2}\frac{rdrd\theta}{z^2 + r^2}\frac{z}{\sqrt{z^2+r^2}} = \frac{1}{4\pi\epsilon_0}\frac{Qz}{\pi R^2}\frac{rdrd\theta}{(z^2+r^2)^{3/2}}\\E_z = \frac{1}{4\pi\epsilon_0}\frac{Qz}{\pi R^2}\int\limits^{2\pi}_0 \int\limits^R_0 {\frac{rdrd\theta}{(z^2+r^2)^{3/2}}} = \frac{1}{4\pi\epsilon_0}\frac{Qz}{\pi R^2} 2\pi(\frac{1}{z} - \frac{1}{\sqrt{z^2+R^2}})$

Substituting the parameters, z = 0.025 m, Q = - 8 x 10^(-9) C, and R = 0.0105 m, yields the final result:

$E_z = \frac{1}{2\epsilon_0}\frac{Qz}{\pi R^2}(\frac{1}{z} - \frac{1}{\sqrt{z^2+R^2}}) = -1.02 \times 10^5~N/C$

(b) We will have a similar approach, but a simpler integral.

$dE = \frac{1}{4\pi\epsilon_0}\frac{dQ}{z^2 + R^2}\\\frac{Q}{2\pi R} = \frac{dQ}{Rd\theta}\\dQ = \frac{Qd\theta}{2\pi}\\dE = \frac{1}{4\pi\epsilon_0}\frac{Qd\theta}{2\pi(z^2 + R^2)}\\dE_z = \frac{1}{4\pi\epsilon_0}\frac{Qd\theta}{2\pi(z^2 + R^2)}\frac{z}{\sqrt{z^2+R^2}} = \frac{1}{4\pi\epsilon_0}\frac{Qzd\theta}{2\pi(z^2 + R^2)^{3/2}}\\E_z = \frac{1}{4\pi\epsilon_0}\frac{Qz}{2\pi(z^2 + R^2)^{3/2}}\int\limits^{2\pi}_0 {} \, d\theta = \frac{1}{4\pi\epsilon_0}\frac{Qz}{2\pi(z^2 + R^2)^{3/2}}2\pi$

$E_z = \frac{1}{4\pi\epsilon_0}\frac{Qz}{(z^2 + R^2)^{3/2}} = -9.07\times 10^4~N/C$

Note that, in this case the source object is a one dimensional hoop rather than a two dimensional disk.

3 0

3 years ago

QUESTION 1

Sveta_85 [38]

distance from the Sun of 2.77 astronomical units or about 414 million km 257 million miles and orbiting period of 4.62 years

7 0

3 years ago