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Rina8888 [55]
3 years ago
9

If wave ripples move across a pond from left to right, in which direction are the water molecules moving?

Physics
2 answers:
umka2103 [35]3 years ago
7 0

The correct answer to the question is that the direction of the vibration of water molecule is perpendicular to the direction of ripple direction. i.e the water molecules will move up and down.

EXPLANATION:

Before going to answer this question, first we have to understand the nature of transverse wave.

A transverse wave is a wave in which the direction of propagation of the wave is perpendicular to the direction of vibration of particles.

As per the question, the ripples are created in a pond and they are moving from left to right.The wave ripples produced in a pond is transverse in nature.

When any body is thrown into the pond, the water particle at the point of disturbance will start vibrating about its mean position, and the process is repeated . The vibrated particles will move up and down about the equilibrium point. But the energy or wave will be transported from particle to particle. Hence, the vibration of particles is normal to wave propagation.

Hence, the water molecules will move up and down when the wave ripples are produced in a pond.

____ [38]3 years ago
4 0
My inference is that the water molecules are moving up and down. I don't know for sure but i'm usually correct.

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Jet aircraft maintenance crews are required to wear protective earplugs. Members of a particular crew wear earplugs that reduce
Taya2010 [7]

Answer:

So the sound intensity level they would experience without the earplugs is 110.32dB.

Explanation:

Given data

Sound intensity by factor =215

Sound intensity level =87 dB

To find

Sound intensity level they would experience without the earplugs

Solution

First we need to find the new sound intensity level

So

I_{n}=215(10^{\frac{87}{10} } )\\I_{n}=1.08*10^{11})

The dB can be calculated as:

dB=10log(I_{n})\\

Substitute the given values

dB=10log(1.08*10^{11})\\dB=110.32dB

So the sound intensity level they would experience without the earplugs is 110.32dB.

7 0
3 years ago
A large plate carries a uniform charge density σ = 8. 85 × 10-9 c/m2. a pattern showing equipotential surfaces with a 5 v potent
vfiekz [6]

The potential difference comes out to be

10 \times 10 {}^{ - 3} m

Given:

σ = 8. 85 × 10-9 c/m2

we know,

E = \frac{σ}{2ε0}

E =  \frac{8.85 \times 10 {}^{ - 9} }{2ε0}

E =  \frac{v}{d}

given the potential difference between two equipotential surface=5v

E=∆v

∆d=∆v/E

=  \frac{5 \times 8.85 \times 10 { }^{ - 12} \times 2 }{8.85 \times 10 {}^{ - 9} }

Δ = 10 \times 10 {}^{ - 3} m

Thus the potential difference is

10 \times 10 {}^{ - 3} m

Learn more about potential difference from here: brainly.com/question/28165869

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5 0
2 years ago
Shayla was late for school. She ran the last half mile in 4 minutes. What was her average speed in miles per hour?
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7.5 miles per  hour

Explanation:

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Margy is trying to improve her cardio endurance by performing an exercise in which she alternates walking and running 100.0 m ea
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Answer:

6.5 m/s

Explanation:

We are given that

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Initial speed, u=1.4 m/s

Acceleration, a=0.20 m/s^2

We have to find the final velocity at the end of the 100.0 m.

We know that

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Using the formula

v^2-(1.4)^2=2\times 0.20\times 100

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3 years ago
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