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3 years ago

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Physics

1 answer:

sergij07 [2.7K]3 years ago

4 0

Based on the nature of constructive interference,

none of the wave intersections will produce constructive interference.

<h3>What is constructive interference?</h3>

Constructive interference is interference that occurs when two waves of the same frequency amplitude and wavelength travelling in the same direction are superposed with the resultant effect of reinforcement of both waves to produce a larger wave.

Constructive interference occurs when the path difference between two identical waves at a point is:

delta s = n × wavelength

where n = 0, 1, 2, ...

For the various intersections:

1.77 cm crest intersecting with a 0.65 cm crest; n = 1.77/0.65 = 2.7
A 1.2 mm crest intersecting with a 3.9 mm trough is destructive.
A 4.55 N trough intersecting with a 1.59 N trough; n = 4.55/1.59 = 2.8
A 0.44 inch trough intersecting with a 0.72 inch crest is destructive.
A 7.42 mm trough intersecting with a 1.93 mm trough; n = 7.42/1.93 = 3.8

Therefore, none of the wave intersections will produce constructive interference

Learn more about constructive interference at: brainly.com/question/1040831

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4 years ago

Read 2 more answers

(a) You short-circuit a 20 volt battery by connecting a short wire from one end of the battery to the other end. If the current

erica [24]

(a) $1.11 \Omega$

When the battery is short-cut, the only resistance in the circuit is the internal resistance of the battery. Therefore, we can apply Ohm's law:

$r=\frac{V}{I}$

where

V = 20 V is the voltage across the internal resistance of the battery

I = 18 A is the current flowing through it

Solving the equation,

$r=\frac{20 V}{18 A}=1.11\Omega$

(b) 360 W

The power generated by the battery is given by the equation

$P=VI$

where

V = 20 V is the voltage of the battery

I = 18 A is the current

Substituting into the formula,

$P=(20 V)(18 A)=360 W$

(c) 360 J

The energy dissipated by the internal resistance is given by

$E=Pt$

where

P = 360 W is the power generated

t = 1 s is the time

Solving the equation, we find

$E=(360 W)(1 s)=360 J$

(d) 1.65 A

The battery is now connected to a $R=11 \Omega$ resistor. This means that the internal resistance of the battery is now connected in series with the other resistor R: so, the total resistance of the circuit is

$R_T = r+R=1.11 \Omega +11 \Omega = 12.11 \Omega$

And so, the current flowing through the circuit is

$I=\frac{V}{R_T}=\frac{20 V}{12.11\Omega}=1.65 A$

(e) 29.9 W

The power dissipated in the external resistor is given by

$P=I^2 R$

where

I = 1.65 A is the current

$R=11 \Omega$ is the resistance

Solving the equation, we find

$P=(1.65 A)^2(11 \Omega)=29.9 W$

(f) 18.17 V

The terminals of the voltmeter are placed at the two end of the battery. The battery provides an emf of 20 V, however due to the internal resistance, some of this voltage is dropped across the internal resistance. Therefore, the actual potential difference that will be read by the voltmeter will be:

$V=\epsilon - Ir =20 V -(1.65 A)(1.11 \Omega)=18.17 V$