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kirza4 [7]
2 years ago
12

Pierce conducts an experiment in which waves collide in a way that the energy increases. What has occurred? refraction reflectio

n constructive interference destructive interference
Physics
2 answers:
Nadya [2.5K]2 years ago
4 0

Answer:

constructive interference.

Explanation:

Vitek1552 [10]2 years ago
3 0

Answer:

constructive interference.

Explanation:

Electromagnetic waves is a propagating medium used in all communications device to transmit data (messages) from the device of the sender to the device of the receiver.

Generally, the most commonly used electromagnetic wave technology in telecommunications is radio waves.

Radio waves can be defined as an electromagnetic wave that has its frequency ranging from 30 GHz to 300 GHz and its wavelength between 1mm and 3000m.

In this scenario, an experiment is conducted by Pierce in which waves collide in a way that the energy increases. Thus, this is an example of constructive interference i.e constructive interference has occurred.

An interference occurs when two pair of waves such as sound and light waves pass through each other i.e crossing each other's path while traveling through the same medium or space.

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An AM radio station broadcasts isotropically (equally in all directions) with an average power of 3.40 kW. A receiving antenna 6
lara [203]

To solve the problem we will apply the concepts related to the Intensity as a function of the power and the area, as well as the electric field as a function of the current, the speed of light and the permeability in free space, as shown below.

The intensity of the wave at the receiver is

I = \frac{P_{avg}}{A}

I = \frac{P_{avg}}{4\pi r^2}

I = \frac{3.4*10^3}{4\pi(4*1609.34)^2} \rightarrow 1mile = 1609.3m

I = 6.529*10^{-6}W/m^2

The amplitude of electric field at the receiver is

I = \frac{E_{max}^2}{2\mu_0 c}

E_{max}= \sqrt{2I\mu_0 c}

The amplitude of induced emf by this signal between the ends of the receiving antenna is

\epsilon_{max} = E_{max} d

\epsilon_{max} = \sqrt{2I \mu_0 cd}

Here,

I = Current

\mu_0 = Permeability at free space

c = Light speed

d = Distance

Replacing,

\epsilon_{max} = \sqrt{2(6.529*10^{-6})(4\pi*10^{-7})(3*10^{8})(60.0*10^{-2})}

\epsilon_{max} = 0.05434V

Thus, the amplitude of induced emf by this signal between the ends of the receiving antenna is 0.0543V

6 0
3 years ago
A bulldozer and a Mini Cooper are involved in a head-on collision. Which one experiences a greater force
Pepsi [2]

Answer:

The mini Cooper will experience the greater force

Explanation:

Generally, a bulldozer has a greater mass compared to a Mini Cooper hence when both of these vehicles interact in an head on collision the Mini Cooper will experience a greater force because the bulldozer has a greater momentum

5 0
2 years ago
An electric dipole consisting of charges of magnitude 1.70 nC separated by 6.80 μm is in an electric field of strength 1160 N/C.
bazaltina [42]

Answer:

p = 1.16 10⁻¹⁴ C m     and  ΔU = 2.7 10 -11 J

Explanation:

The dipole moment of a dipole is the product of charges by distance

                        p = 2 a q

With 2a the distance between the charges and the magnitude of the charges

                        p = 1.7 10⁻⁹ 6.8 10⁻⁶

                        p = 1.16 10⁻¹⁴ C m

 

The potential energie dipole  is described by the expression

                       U = - p E cos θ

Where θ is the angle between the dipole and the electric field, the zero value of the potential energy is located for when the dipole is perpendicular to the electric field line

Orientation parallel to the field

                      θ = 0º

                      U = 1.16 10⁻¹⁴ 1160 cos 0

                      U1 = 1.35 10⁻¹¹ J

Antiparallel orientation

                       θ = 180º

                      cos 180 = -1

                      U2 = -1.35 10⁻¹¹ J

The difference in energy between these two configurations is the subtraction of the energies

                         ΔU = | U1 -U2 |

                         ΔU = 1.35 10-11 - (-1.35 10-11)

                         ΔU = 2.7 10 -11 J

6 0
3 years ago
The ratio of the speed of light in a medium to the speed of light in a vacuum
Liula [17]

Answer:D.Refractive Indez

Explanation:

It is usually expressed the other way: the ratio of the speed of light in a vacuum to the speed of light in a medium. In that case, it is called the "index of refraction".

7 0
3 years ago
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The surface area of a postage
ANTONII [103]

Answer:

608

Explanation:

Trust me

4 0
2 years ago
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