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kirza4 [7]
3 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]3 years ago
4 0

Answer:

constructive interference.

Explanation:

Vitek1552 [10]3 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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A toroidal coil has a mean radius of 16 cm and a cross-sectional area of 0.25 cm2; it is wound uniformly with 1000 turns. A seco
Roman55 [17]

Answer:

Explanation:

Mutual inductance is equal to magnetic flux induced in the secondary coli due to unit current in the primary coil .

magnetic field in a torroid  B = μ₀ n I , n is number of turns per unit length and I is current .

B = 4π x 10⁻⁷ x (1000 / 2π x .16  )x 1 ( current = 1 A)

flux in the secondary coil

= B x area of face of coil x no of turns of secondary

= 4π x 10⁻⁷ x (1000 /2π x .16  ) .25 x 10⁻⁴ x 750

= 2 x 1000 x .25 x( 750 /.16) x 10⁻¹¹

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6 0
3 years ago
Read 2 more answers
At a certain distance from a point charge, the potential and electric field magnitude due to that charge are 4.98 V and 12.0 V/m
Finger [1]

Answer:

1. d = 0.415 m.

2. Q = 2.285 x 10^{-10} C.

Explanation:

The electric field and potential can be found by the following equations:

E = \frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}\\V = \frac{1}{4\pi\epsilon_0}\frac{Q}{r}

Applying these equations to the given variables yields

E = 12 = \frac{1}{4\pi\epsilon_0}\frac{Q}{d^2}\\V = 4.98 = \frac{1}{4\pi\epsilon_0}\frac{Q}{d}

Divide the first line to the second line:

\frac{12}{4.98} = \frac{ \frac{1}{4\pi\epsilon_0}\frac{Q}{d^2}}{\frac{1}{4\pi\epsilon_0}\frac{Q}{d}}\\\frac{12}{4.98} = \frac{1}{d}\\d = 0.415~m

Using this distance in either of the equations give the magnitude of the charge.

12 = \frac{1}{4\pi\epsilon_0}\frac{Q}{(0.415)^2}\\12 = \frac{1}{4\pi (8.8\times 10^{-12})}\frac{Q}{(0.415)^2}\\Q = 2.285 \times 10^{-10}~C

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3 years ago
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Answer:

Explanation:

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So total electric flux through closed surface will be zero.  

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(a) The electric field is zero everywhere on the surface.  True  

(b) The electric field is normal to the surface everywhere on the surface. - false

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