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nalin [4]
3 years ago
10

An aurora occurs when?

Physics
1 answer:
Vinil7 [7]3 years ago
4 0

energetic electrically charged particles (mostly electrons) accelerate along the magnetic field lines into the upper atmosphere, where they collide with gas atoms, causing the atoms to give off light.

(exploratorium.com)

gotta cite those sources

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Pls help i begg youuuuu
Anna [14]
Pitch is the sensation of certain frequencies to the ear. High frequency = high pitch, low frequency = low pitch. 

f = c(speed of the wave) /  <span>λ (wavelength)

1. 343m/s / 0.77955m = 439.99 Hz   
     This corresponds to pitch A 

2. 343m/s / 0.52028m = 659.26 Hz
</span>     This corresponds to pitch E 
<span>
3. 343m/s / 0.65552m = 523.349 Hz
    </span>This corresponds to pitch C

4. using f = c /  λ
  λ = c / f<span>
     = 343m/s / 587.33 = 0.583999 m = 0.584 m

</span>
3 0
3 years ago
The electric potential inside a charged spherical conductor of radius R is is given by V = keQ/R, and the potential outside is g
Bumek [7]

Answer:

For outer points of shell

E = \frac{k_eQ}{r^2}

Now for inner point of shell

E = 0

Explanation:

As we know that out side the shell electric potential is given as

V = \frac{K_e Q}{r}

inside the shell the electric potential is given as

V = \frac{K_e Q}{R}

now we know the relation between electric potential and electric field as

E = - \frac{dV}{dr}

so we can say for outer points of the shell

E = -\frac{dV}{dr}

E = - \frac{d}{dr}(\frac{K_eQ}{r})

E = \frac{k_eQ}{r^2}

Now for inner point again we can use the same

E = - \frac{dV}{dr}

E = - \frac{d}{dr}(\frac{K_eQ}{R})

E = 0

6 0
3 years ago
The Earth's magnetic field is approximately 0.5 Gauss. What is the magnetic energy density of this field
ser-zykov [4K]

Answer:

The  magnetic energy density  is  E_B  =  9.95 *10^{-4} \  J/m^3

Explanation:

From the question we are told that  

    The  earths magnetic field is  B  =  0.5 \  Gauss

Now

          1 \ Gauss  \to  1 *10^{-4} \ T

=>        0.5  \ Gauss  \to  x T

So  

          x =  \frac{0.5 *  1*10^{-4}}{1}

          x = B= 5.0 *10^{-5} \  T

Generally the magnetic energy density is  mathematically represented as

        E_B  =  \frac{B^2}{ 2 *  \mu_o  }

  Here \mu_o is the  permeability of  free space with a constant value

          \mu_o  =  4\pi * 10^{-7} N/A^2

substituting values for equation above  

       E_B  =  \frac{(5*10^{-5})^2}{ 2 *   4\pi * 10^{-7}  }

       E_B  =  9.95 *10^{-4} \  J/m^3

5 0
3 years ago
Which of the following statements are true?
Tems11 [23]

Answer:

1) A time-varying magnetic field will produce an electric field.

4) A time-varying electric field will produce a magnetic field.

Explanation:

1) A time-varying magnetic field will produce an electric field.

TRUE

time varying magnetic field will produce electric field which is given as

E = \frac{r}{2}\frac{dB}{dt}

2) Time-varying electric and magnetic fields can propagate through space only if there is no matter in their path.

FALSE

Time varying electric field and magnetic field will induce each other and it can travel through any medium as well as it can travel without any medium also

3) Electric and magnetic fields can be treated independently only if they vary in time.

FALSE

electric field can be due to stationary charge and magnetic field due to current carrying elements so it is not compulsory to have time varying

4) A time-varying electric field will produce a magnetic field.

TRUE

Time varying electric field will produce magnetic field given as

B = \frac{\mu_o\epsilon_o A}{2r}\frac{dE}{dt}

3 0
3 years ago
A 10-kg object is dropped from rest. after falling a distance of 50 m, it has a speed of 26 m/s. what is the change in mechanica
likoan [24]

The change in mechanical energy caused by the dissipative resistance force is equal to, difference between the potential energy and kinetic energy of the object.

Potential energy of the object, P.E = mgh

m is mass of the object = 10 kg

g is acceleration due to gravity = 9.8 m/s²

h= height from which it is dropped =50 m

Substituting the value we get,

P.E = 10×9.8×50 = 4900 J

Kinetic energy of the object, K.E = \frac{1}{2}mv^{2}

v is the velocity of the object = 26 m/s²

K.E = (1/2)×10×(26)²

= 3380 J

Change in mechanical energy caused by dissipative force = P.E ₋ K.E

= 4900 ₋ 3380 = 1520 J

4 0
3 years ago
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