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Alex Ar [27]
3 years ago
13

if two waves with equal amplitudes and wavelengths travel through a medium in such a way that a particular particle of the mediu

m is at the crest of one wave and at the trough of the other wave at the same time, what will happen to that particle?
Physics
2 answers:
Novosadov [1.4K]3 years ago
8 0
If two waves with equal amplitude and wavelength travel through a medium such that particular particle is at the crest of one wave and at the trough of another wave, then destructive interference will take place.Both waves will cancel each other out at that point such that a node will form in which the particle will remain at its rest position only and will not move. 
Irina-Kira [14]3 years ago
8 0

Answer:

As per the law of superposition we know that when two waves superimpose at a point on the medium then the resultant displacement of the medium particle is vector sum of the displacement due to each wave.

So here by superposition principle we can write

y_{net} = y_1 + y_2

now here we know that one of the wave reaches with crest while other wave reaches at trough position.

So we will have

y_1 = +A

y_2 = -A

so here by superposition we will have

y_{net} = A - A = 0

so after superposition the medium particle will come to its mean position.

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Snezhnost [94]

Explanation:

Let N be the number of turns in a circular loop having a radius of r₀ and a resistance R. The magnetic field is given by :

B(t)=B_oe^{-\dfrac{t}{\tau}}

The induced emf in the circular coil is given by :

\epsilon=\dfrac{-d\phi}{dt}

Where

\phi=BA

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\epsilon=\dfrac{NAB_o}{\tau}e^{-t/\tau}

Power dissipated is given by :

P=\dfrac{\epsilon^2}{R}

P=\dfrac{(\dfrac{NAB_o}{\tau}e^{-t/\tau})^2}{R}

P=(\dfrac{NAB_o}{\tau})^2\dfrac{e^{-2t/\tau}}{R}

Energy dissipated in the circuit is given by :

E=\int\limits^T_0 {P.dt}

E=\int\limits^T_0 {(\dfrac{NAB_o}{\tau})^2\dfrac{e^{-2t/\tau}}{R}.dt}

E=\dfrac{1}{R}(\dfrac{NAB_o}{\tau})^2\int\limits^T_0 {e^{-2t/\tau}.dt}

E=\dfrac{(N\pi r_o^2B_o^2)^2}{2R}{\tau(e^{2t/\tau}-1)}{}

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7 0
3 years ago
if change in blood pressure between the brain and the feet is 1.88×10^4 pa .what will be the volume flow rate from head to feet
klio [65]

Answer: 3765.66 \frac{m^{3}}{s}

Explanation:

We can solve this problem using the <u>Poiseuille equation</u>:

Q=\frac{\pi r^{4}\Delta P}{8\eta L}

Where:

Q  is the Volume flow rate

r=23 cm \frac{1 m}{100 cm}=0.23 m  is the effective radius

L=6 ft \frac{0.3048 m}{1 ft}=1.8288 m  is the length

\Delta P=1.88(10)^{4} Pa  is the difference in pressure

\eta=3(10)^{-3} Pa.s is the viscosity of blood

Solving:

Q=\frac{\pi (0.23 m)^{4}(1.88(10)^{4} Pa)}{8(3(10)^{-3} Pa.s)(1.8288 m)}

Q=3765.66 \frac{m^{3}}{s}

4 0
4 years ago
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its A
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