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DENIUS [597]
3 years ago
11

Two waves with amplitudes of 75 units and 74 units arrive at a point in a medium simultaneously. If the two waves are out of pha

se with each other, what is the resultant amplitude of the wave?
a. 74 units

b. 1 unit

c. 75 units

d. m149 units
Physics
2 answers:
liraira [26]3 years ago
7 0
<h3><u>Answer;</u></h3>

1 unit

<h3><u>Explanation;</u></h3>
  • <em><u>Interference of a wave</u></em> is the property a wave that occurs when two waves travelling in the same medium meet at a point or superpose at a point in the medium.
  • <em><u>Interference may be constructive or destructive. Constructive occurs when two identical waves travelling in the same medium in phase meet at a point.</u></em> The resulting wave has twice the amplitude of the individual waves.
  • <em><u>Destructive interference occurs when two waves travelling in the same medium meet and out of phase.</u></em> The resulting amplitude will be zero if the initial waves are identical, or the difference between the amplitude of the original waves if they are not identical.
  • Therefore; <em><u> If the two waves one with an amplitude 75 units and the other 74 units, are out of phase with each other, the resultant amplitude of the resulting wave will be 1 unit, the difference between the amplitude of the original waves.</u></em>
avanturin [10]3 years ago
6 0
If the two waves have the SAME FREQUENCY and are exactly
out of phase (180° apart), then the resultant wave will have the
same frequency and an amplitude of 1 unit.

If the two waves do not have the SAME FREQUENCY, then their
relative phase is meaningless.
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Brayden and Riku now use their skills to work a problem. Find the equivalent resistance, the current supplied by the battery and
Liono4ka [1.6K]

a) 5 \Omega, 1.6 A

b) 6 \Omega, 1.33 A

Explanation:

a)

In this situation, we have two resistors connected in series.

The equivalent resistance of resistors in series is equal to the sum of the individual resistances, so in this circuit:

R=R_1+R_2

where

R_1=4\Omega

R_2=1 \Omega

Therefore, the equivalent resistance is

R=4+1=5 \Omega

Now we can use Ohm's Law to find the current flowing through the circuit:

I=\frac{V}{R}

where

V = 8 V is the voltage supplied by the battery

R=5\Omega is the equivalent resistance of the circuit

Substituting,

I=\frac{8}{5}=1.6 A

The two resistors are connected in series, therefore the current flowing through each resistor is the same, 1.6 A.

b)

In this part, a third resistor is added in series to the circuit; so the new equivalent resistance of the circuit is

R=R_1+R_2+R_3

where:

R_1=4\Omega\\R_2=1\Omega\\R_3=1\Omega

Substituting, we find the equivalent resistance:

R=4+1+1=6 \Omega

Now we can find the current through the circuit by using again Ohm's Law:

I=\frac{V}{R}

where

V = 8 V is the voltage supplied by the battery

R=6\Omega is the equivalent resistance

Substituting,

I=\frac{8}{6}=1.33 A

And the three resistors are connected in series, therefore the current flowing through each resistor is the same, 1.33 A.

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3 years ago
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