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3 years ago

What will occur when the trough of wave A overlaps the trough of wave B?

Physics

2 answers:

olganol [36]3 years ago

5 0

Answer:

Constructive interference

Explanation:

The superposition of two wave is called interference phenomenon. After interference of wave, the amplitude of resultant wave can be more, less or remains the same.

The two conditions of interference are :

1. Constructive interference 2. Destructive interference

According to given question, when the trough of wave A overlaps the trough of wave B constructive interference occurs. The amplitude of resultant wave is greater than the individual waves. Also, the energy of the resultant wave is more than the energies of wave A and wave B.

densk [106]3 years ago

3 0

Constructive interference will occur, which means the waves will combine.

In destructive inference, the waves cancel each other out.

Hope this helps :)

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What's unique about the philosophical approach to the question of truth?

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Answer:

It requires evidence or proof and to avail something is true.

Explanation:

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3 years ago

A parallel-plate vacuum capacitor has 8.60 J of energy stored in it. The separation between the plates is 3.80 mm . If the separ

creativ13 [48]

Answer:

Part a)

$E = \frac{8.60}{2.62} = 3.28 J$

Part b)

$E = 2.62(8.60) = 22.5 J$

Explanation:

As we know that the energy of capacitor when it is not connected to potential source is given as

$U = \frac{Q^2}{2C}$

As we know that initial energy is given as

$8.60 = \frac{Q^2}{2C}$

now we know that capacitance of parallel plate capacitor is given as

$C = \frac{\epsilon_0A}{d}$

now the new capacitance when distance is changed from 3.80 mm to 1.45 mm

$C' = \frac{Cd}{d'}$

$C' = \frac{C(3.80)}{1.45}$

$C' = 2.62 C$

Now the new energy of the capacitor is given as

$E = \frac{Q^2}{2(2.62C)}$

$E = \frac{8.60}{2.62} = 3.28 J$

Part b)

Now if the voltage difference between the plates of capacitor is given constant

now the energy energy of capacitor is

$U = \frac{1}{2}CV^2$

$8.60 = \frac{1}{2}CV^2$

now when capacitance is changed to new value then new energy is given as

$E = \frac{1}{2}C'V^2$

$E = \frac{1}{2}(2.62C)V^2$

$E = 2.62(8.60) = 22.5 J$

6 0

3 years ago

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