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lianna [129]
3 years ago
14

Suppose that the electric field in the Earth's atmosphere is E = 8.60 101 N/C, pointing downward. Determine the electric charge

in the Earth. (The radius of the Earth is 6371 km, and the Coulomb's constant, ke, is 8.99 109 N · m2/C2.)
Physics
1 answer:
asambeis [7]3 years ago
4 0

Answer:

q = 3.87 x 10⁵ C

Explanation:

given,

Electric field, E = 8.60 x 10¹ = 86 N/C

radius of earth, R = 6371 Km = 6.371 x 10⁶ m

Coulomb constant, K = 9 x 10⁹ N · m²/C²

Charge on the earth = ?

the electric field at the point

E =\dfrac{kq}{r^2}

q =\dfrac{Er^2}{k}

inserting all the values

q =\dfrac{86\times (6.371\times 10^6)^2}{9\times 10^{9}}

      q = 3.87 x 10⁵ C

The electric charge on the earth is equal to 3.87 x 10⁵ C

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A grating has 2000 slits/cm/cm. How many full spectral orders can be seen (400 to 700 nmnm) when it is illuminated by white ligh
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With the use of the formula SinФ = nλ / d, there are 16 spectral orders which can be seen when it is illuminated by white light.

Given that a grating has 2000 slits/cm. That is,

d = 0.01 / 2000

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The wavelength λ = (700 - 400) nm

λ = 300 x 10^{-9} m

To calculate how many full spectral orders that can be seen (400 to 700 nm) when it is illuminated by white light, we will use the below formula

SinФ = nλ / d

Φ = Sin^{-1} (nλ / d)

When n = 1

Φ = Sin^{-1}(300 x 10^{-9}/ 5 x 10^{-6} )

Φ = 3.4 degrees

when n = 2

Φ = Sin^{-1}(2 x 300 x 10^{-9}/ 5 x 10^{-6} )

Ф = 6.9 degrees

When n = 3

Ф = Sin^{-1}(3 x 300 x 10^{-9}/ 5 x 10^{-6} )

When n = 16

Ф = Sin^{-1}(16 x 300 x 10^{-9}/ 5 x 10^{-6} )

Ф = Sin^{-1}(0.96)

Ф = 73.7 degrees

when n = 17

Ф = Sin^{-1}(17 x 300 x 10^{-9}/ 5 x 10^{-6} )

Ф = Sin^{-1}(1.05)

Ф = Error ( that is, it does not exist)

Therefore, there are 16 spectral orders which can be seen when it is illuminated by white light.

Learn more about double slit here: brainly.com/question/4449144

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Answer: It would be 12 m/s.

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\alpha =\frac{2*\pi *t}{24H}  t being expressed in hours.

\alpha =\frac{2*\pi *11.9s*1H/3600s}{24H}=0.000865rad

From the triangle formed by A,B and the center of the planet, we know that:

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