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Mashcka [7]
3 years ago
9

Does changing position of charges change the magnitude

Physics
1 answer:
hodyreva [135]3 years ago
7 0

Answer:

No, they will not change.

Explanation:

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At some instant and location, the electric field associated with an electromagnetic wave in vacuum has the strength 71.9 V/m. Fi
kirza4 [7]

Answer:

a) Magnetic field strength, B = 2.397 * 10⁻⁷ T

b) Total energy density, U = 4.58 * 10⁻⁸ J/m³

c) Power flow per unit area, S = 13.71 W/m²

Explanation:

a) Electric field strength, E = 71.9 V/m

The relationship between the Electric field strength and the magnetic field strength in vacuum is:

E = Bc where c = 3.0 * 10⁸ m/s

71.9 = B * 3.0 * 10⁸

B = 71.9 / (3.0 * 10⁸)

B = 23.97 * 10⁻⁸

B = 2.397 * 10⁻⁷ T

b) Total Energy Density:

U = \frac{1}{2} \epsilon_0E^2 +  \frac{1}{2} \frac{B^2}{\mu_0} \\U = \frac{1}{2}* 8.85 * 10^{-12}*71.9^2 +  \frac{1}{2} \frac{(2.397*10^{-7})^2}{4\pi*10^{-7}}\\U = 2.29 * 10^{-8} + 2.29 * 10^{-8}\\U = 4.58 * 10^{-8} J/m^3

c)Power flow per unit area

S = \frac{1}{\mu_0} EB\\S = \frac{1}{4\pi * 10^{-7} } * 71.9 * 2.397 * 10^{-7}\\S = 13.71 W/m^2

6 0
3 years ago
Take schlatts love uwu (i cant spell)
7nadin3 [17]

thank you so much for the schlatt

8 0
3 years ago
A clothes dryer in a home draws a current of 10 amps when connected on a special 220-volts household circuit.what is the resista
aliya0001 [1]

Answer:

22Ω

Explanation:

if    V ⇒ voltage

      I ⇒ current

      R ⇒ resistance

V = IR

220 = 10 x R

220 / 10 = R

22 = R

8 0
2 years ago
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diamong [38]

Answer:

tectonic plate movement

Explanation:

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2 years ago
What is the greatest distance an image can be located behind a convex spherical mirror?
Afina-wow [57]

Answer:

Maximum distance of image from mirror is equal to focal length of the mirror

Explanation:

As we know by the equation of mirror we have

\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}

here we know for convex mirror

object position is always negative as it will be placed behind the mirror always

while the focal length of the convex mirror is always taken positive

So here we have

\frac{1}{d_i} + \frac{1}{-d_o} = \frac{1}{f}

\frac{1}{d_i} = \frac{1}{d_o} + \frac{1}{f}

so here maximum value of image distance is equal to focal length of the mirror

6 0
3 years ago
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