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-Dominant- [34]
3 years ago
8

What best describes the angle between a changing electric field and the electromagnetic wave produced by it? (2 points)

Physics
2 answers:
Leno4ka [110]3 years ago
6 0

Answer:

Option (A) is the correct answer of this question.

Explanation:

The angle between a changing electric field and the induced electromagnetic wave is always equal to the right angle. The electrical field, the magnetic field and the EM wave are perpendicular. This gives its transverse form to the electromagnetic waves.

  • A shifting electric field generates a changing magnetic field that in effect causes a changing electrical field.
  • This means the source has generated a perpendicular to each other, shifting electrical and magnetic fields that travel away from the source.
  • Electromagnetic (EM) waves shift electrical and magnetic fields, conveying energy and momentum across space.

According to this question, other options are incorrect.

Oksanka [162]3 years ago
4 0

A) always equal to a right angle

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A Ping-Pong ball with mass 2.5 g is attached by a thread to the bottom of a beaker. When the beaker is filled with water so that
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0.022m or 2.2cm

Expxlanation:

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Data obtained from the question. This includes:

Mass (m) = 2.5g = 2.5/1000 = 2.5x10^-3Kg

Tension (T) = 0.029 N

Density (ρ) = 1000 kg/m3

Acceleration due to gravity (g) = 9.81 m/s2

Diameter (d) =?

Step 2:

Finding an expression to calculate the diameter of the ball. This is illustrated below:

Tension = weight displaced - weight of the ball

Weight displaced = Mass of water x acceleration due to gravity

Mass of water = Density x volume

Mass of water = ρxV

Weight displaced = ρxVxg = ρVg

Weight of the ball = Mass of the ball x acceleration due to gravity

Weight of the ball = mg

Therefore,

Tension = weight displaced - weight of the ball

T = ρVg - mg

Make V the subject of the formula

T = ρVg - mg

T + mg = ρVg

Divide both side by ρg

V = ( T + mg) /ρg. (1)

Recall that the ball is spherical in shape and the Volume of a sphere is given by

V = 4/3πr^3

Radius (r) = diameter (d) /2

V = 4/3π(d/2)^3

V = 4/3πd^3/8

V = πd^3 /6

Substituting the value of V into equation 1, we have

V = ( T + mg) /ρg

πd^3 /6 = ( T + mg) /ρg.

Making d the subject of the formula, we have:

πd^3 /6 = (T + mg) /ρg.

d^3 = 6(T + mg) /πρg.

Taking the cube root of both sides

d = [6(T + mg) /πρg]^1/3

Step 3:

Determination of the diameter of the ball. This is illustrated below:

T = 0.029 N

m = 2.5x10^-3Kg

g = 9.81 m/s2

ρ = 1000 kg/m3

d =?

d = [6(T + mg) /πρg]^1/3

d = [6(0.029 + 2.5x10^-3x9.81)/ πx1000x9.81]^1/3

d = 0.022m

Therefore, the diameter of the ball is 0.022m or 2.2cm

6 0
3 years ago
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