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sergeinik [125]
3 years ago
10

The earth has a vertical electric field at the surface, pointing down, that averages 100 N/C. This field is maintained by variou

s atmospheric processes, including lightning. If research came out on Planet x in a distant solar system that had a electric field with strength 222 N/C and 0.6 the radius of the earth, what would be the excess charge on planet x
Physics
1 answer:
zlopas [31]3 years ago
4 0

Answer:

q = 3.6 10⁵  C

Explanation:

To solve this exercise, let's use one of the consequences of Gauss's law, that all the charge on a body can be considered at its center, therefore we calculate the electric field on the surface of a sphere with the radius of the Earth

          r = 6 , 37 106 m

          E = k q / r²

          q = E r² / k

          q = \frac{100 \ (6.37 \ 10^6)^2}{9 \ 10^9}

          q = 4.5 10⁵ C

Now let's calculate the charge on the planet with E = 222 N / c and radius

           r = 0.6 r_ Earth

           r = 0.6 6.37 10⁶ = 3.822 10⁶ m

           E = k q / r²

            q = E r² / k

            q = \frac{222 (3.822 \ 10^6)^2}{ 9 \ 10^9}

            q = 3.6 10⁵  C

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MakcuM [25]
In an atom, you have three particles: protons, electrons, and neutrons. 

When you add or subtract electrons from an atom, it means that there are either more protons than electrons (positively charged) or there are more electrons than protons (negatively charged).

You cannot add or subtract protons since they are the atomic number, which means the number of protons determines the element. Neutrons have a neutral charge, so they wouldn't affect the charge at all.
8 0
3 years ago
At a maximum level of loudness, the power output of a 75-piece orchestra radiated as sound is 72.0 W. What is the intensity of t
FrozenT [24]

Answer:

<em> The intensity would be 0.01432 W/</em>m^{2}<em></em>

Explanation:

The sound intensity is the power that is transmitted by a sound wave per unit area.

Given that

Power output P of the 75-piece orchestra = 72.0 W

The listener's distance  r = 20.0 m

The intensity of sound waves (I) can be obtained with the expression below;

I = P/A .............. 1

where P is the power

A is the area, in this case, at the distance r the sound is radiated through  a sphere.

A = area of a sphere = 4πr^{2}

putting it into equation 1, the intensity would be;

I = P/ 4πr^{2}

Substituting our values we have;

I = 72.0 W /  4 π 20^{2} m^{2}

I = 0.01432 W/m^{2}

Therefore the intensity would be 0.01432 W/m^{2}

3 0
3 years ago
A 4 kg object is moving at a speed of 5 m/sec. How much kinetic energy does the object have? A. 10 joules B. 20 joules C. 50 jou
sp2606 [1]

Answer:

A. 10 joules

Explanation:

Given parameters:

Mass of object  = 4kg

Speed  = 5m/s

Unknown:

Kinetic energy of the the object  = ?

Solution:

Kinetic energy is the energy due to a moving body.

So;

           mathematically;

    K.E  = \frac{1}{2} x m v²

m is the mass

v is the velocity

  K.E  =  \frac{1}{2}  x 4 x 5  = 10J

5 0
3 years ago
Balanced or un balanced
sattari [20]

Answer: unbalance

Explanation:

5 0
3 years ago
A system consists of N particles that can occupy two energy levels: a nondegenerate ground state and a three-fold degenerate exc
cupoosta [38]

Answer:

Ng = 0.893 N,  Ne = 0.107N

Explanation:

Number of particles in Ground state = Ng

Number of particles in Excited state = Ne

Ne/Ng = e^{(-ΔE)/kt}

Since excited state is 3 fold degenerate

Ne/Ng =3 x e^{(-ΔE)/kt}

ΔE = Energy difference between ground and excited states = 0.25eV

T = 960 K

Constant k = 8.617 x 10^-5 eV/K

Ne/Ng = 3 x e^{-0.25/(8.617x10^-5) x 960}

           = 3 x e^(-3.188645)

           = 3 x 0.0412 = 0.1237 ≅ 0.12

Ne = 0.12 Ng

but Ne + Ng = N, where is N is total number of particles, substituting Ne into equation we get,

Ng(1 + 0.12) = N

Ng = N/1.12 = 0.893N

and Ne = 0.12 x 0.893 N = 0.107 N

3 0
3 years ago
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