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3 years ago

What do ocean waves and sound waves have in common?

2 answers:

7 0

D. Both exhibit the same particle-to-particle interaction.Because disturbance is propagated with the help of particles. Other than this,[ light waves are electromagnetic waves. ocean waves and sound waves are mechanical waves. they are able to transfer energy. electromagnetic wave and ocean waves are transverse waves while sound waves are the longitudinal wave. they show wave properties: reflection, refraction, diffraction, interference, and plane-polarization. longitudinal waves such as sound waves cannot be plane-polarized]. The one in the box shows different examples of waves with their examples. Hope it helps.

Nataly [62]3 years ago

4 0

Answer:

E. Both are mechanical waves.

Explanation:

As we know that sounds waves are longitudinal waves in which medium particles will oscillates parallel to the wave propagation.

While in ocean the waves are of both nature i.e. transverse and longitudinal both. Here the energy is transferred to all molecules inside the water by transverse nature of wave while in the case of surface the energy is transferred by longitudinal waves.

So here we can say that common thing between them is that both type of waves required medium particles to propagate energy through the medium.

So all those waves which required medium particles to propagate energy then it is known as mechanical waves

So here correct answer will be

E. Both are mechanical waves.

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4 years ago

You are sitting 3 m away from you friend who is watching a cartoon on his phone. How will the sound itensity change if your frie

Zinaida [17]

Answer:

Decreases by $3.6 \times 10^{-3}$ times

Explanation:

The intensity of a sound is defined as the energy of the sound that is flowing in an unit time through the unit area which is in the direction that is perpendicular to the direction of the sound waves movement.

The intensity of energy is described by the inverse square law. It states that the intensity varies inversely with the distance square of the distance.

In other words, the sound intensity decreases as inversely proportional to the squared of the distance. i.e. $\frac{1}{r^2}$

In the context when the distance was 3 m, the intensity of the sound was = $\frac{1}{9}$

But when the distance became 6 cm or 0.06 m, the sound intensity decreases by = $\frac{1}{0.06^2}$

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3 years ago

A potential difference of 300 volts is applied to a 2.0-µf capacitor and an 8.0-µf capacitor connected in series. (a) What are t

zubka84 [21]

Answer:

a) Q1=Q2=480μC V1=240V V2=60V

b) Q1=96μC Q2=384μC V1=V2=48V

c) Q1=Q2=0C V1=V2=0V

Explanation:

Let C1 = 2μC and C2=8μC

For part (a) of this problem, we know that charge in a series circuit, is the same in C1 and C2. Having this in mind, we can calculate equivalent capacitance first:

$C=\frac{1}{\frac{1}{C1} +\frac{1}{C2} } = 1.6\mu C$

$Q_{T} = V_{T}*C_{T} = 480\mu C$

$V1 = \frac{Q1}{C1}=240V$

$V2 = \frac{Q2}{C2}=60V$

For part (b), the capacitors are in parallel now. In this condition, the voltage is the same for both capacitors:

$V1=V2$ So, $\frac{Q1}{C1} = \frac{Q2}{C2}$

Total charge is the same calculated for part (a), so:

$\frac{Qt - Q2}{C1} = \frac{Q2}{C2}$ Solving for Q2:

Q2 = 384μC Q1 = 96μC.

Therefore:

V1=V2=48V

For part (c), both capacitors would discharge, since their total voltage of 300V would by applied to a wire (R=0Ω). There would flow a huge amount of current for a short period of time, and capacitors would be completely discharged. Q1=Q2=0C V1=V2=0V

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3 years ago