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matrenka [14]
2 years ago
8

Write equations for both the electric and magnetic fields for an electromagnetic wave in the red part of the visible spectrum th

at has a wavelength of 725 nm and a peak electric field magnitude of 2. 9 V/m. (Use the following as necessary: t and x. Assume that E is in volts per meter, B is in teslas, t is in seconds, and x is in meters. Do not include units in your answer. Assume that E
Physics
1 answer:
elena-14-01-66 [18.8K]2 years ago
8 0

The peak magnetic field of the electromagnetic wave in the red part of the visible spectrum is 9.67 x 10⁻¹⁰ T.

<h3>Relationship between electric and magnetic field</h3>

The relationship between electric and magnetic field at a given peak electric field is given as;

c = (E₀) / (B₀)

where;

  • c is speed of light
  • E₀ is the peak electric field
  • B₀ is the peak magnetic field

B₀ = E₀ / c

B₀ = (2.9) / (3 x 10⁹)

B₀ = 9.67 x 10⁻¹⁰ T

Thus, the peak magnetic field of the electromagnetic wave in the red part of the visible spectrum is 9.67 x 10⁻¹⁰ T.

Learn more about peak magnetic field here: brainly.com/question/24487261

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EMERGENCY! PLEASE HELP!
hichkok12 [17]

Answer: Volume = 1080m^3

Explanation:

Given that the prism has a 15 m by 18 m rectangular base and a height of 4 m

Volume is the product of length, breath and height. That is

Volume = L × B × H

Where

L = 18 m

B = 15m

H = 4m

Using the formula above gives:

Volume V = 18 × 15 × 4

V = 1080 m^3

8 0
3 years ago
Which question cannot be answered through making measurements?
Margarita [4]

Answer: it would be A

Explanation: how are we to measure the air of a square mile

8 0
3 years ago
A bus travels east for 3 km the north for 4 km what is its final displacement
lesya [120]
Im about to the math for this right now.
8 0
3 years ago
(a) In what direction would the ship in Exercise 3.57 have to travel in order to have a velocity straight north relative to the
shtirl [24]

Answer:

Direction of ship: 9.45° West of North

Ship's relative speed: 7.87m/s

Explanation:

A. Direction of ship: since horizontal of the velocity of boat relative to the ground is 0

Vx=0

Therefore, -VsSin∅+VcCos∅40°

Sin∅ = Vc/Vs × Cos 40°

Sin∅ = 1.5/7 ×Cos40°

Sin∅= 0.164

∅= Sin-¹ (0.164)

∅= 9.45° W of N

B. Ship's relative speed:

Vy= VsCos∅ + Vcsin40°

= 7Cos9.45° + 1.5sin40°

= 7×0.986 + 1.5×0.642

= 7.865

= 7.87m/s

4 0
3 years ago
Unless indicated otherwise, assume the speed of sound in air to be v = 344 m/s. You have a stopped pipe of adjustable length clo
faltersainse [42]

Answer:

Length of pipe = 0.057 meter

Explanation:

Speed of a transverse wave on a string

v = \sqrt{\frac{F}{\mu} }

where F is the tension in string and \mu is the mass per unit length

Thus,

\mu = \frac{m}{L}

Substituting the given values we get -

\mu = \frac{7.25 * 10^{-3}}{0.62}\\mu = 0.0117 \frac{Kg}{m}

Speed of a transverse wave on a string

v = \sqrt{\frac{4510}{0.0117} } \\v = 620.86 \frac{m}{s}

For third harmonic wave , frequency is equal to

f = \frac{nv}{2L}

Substituting the given values, we get -

f = \frac{3 * 620.86}{2 * 0.62} \\f = 1502.08

Length of pipe

L = \frac{nv}{4 f}

Substituting the given values we get

n = 1 for first harmonic wave

L = \frac{344* 1}{4*1502.08} \\L = 0.057

Length of pipe = 0.057 meter

5 0
3 years ago
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