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liubo4ka [24]
2 years ago
7

The distance between the crest and trough of an ocean wave is 1 meter. What is the amplitude of the wave?​

Physics
1 answer:
Darya [45]2 years ago
8 0

Answer:

distance between crust and trough = 1m

amplitude = 1/2 = 0.5

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A Explain why the sound produced by the horn of an approaching car
aleksklad [387]
When a car approaches you, the sound waves that reach you have a shorter wavelength and a higher frequency. You hear a sound with a higher pitch. When the car moves away from you, the sound waves that reach you have a longer wavelength and lower frequency.

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An approaching source moves closer during period of the sound wave so the effective wavelength is shortened, giving a higher pitch since the velocity of the wave is unchanged. Similarly the pitch of a receding sound source will be lowered.

The Doppler effect is an effect observed in light and sound waves as they move toward or away from an observer. One simple example of the Doppler effect is the sound of an automobile horn. Picture a person standing on a street corner. A car approaches, blowing its horn.

Comparing two waves of the same wavelength, a higher frequency is associated with faster movement. Comparing two waves of different wavelengths, a higher frequency doesn't always indicate faster movement, although it can. Waves of different wavelengths can have the same frequency.

The pitch of a sound is our ear's response to the frequency of sound. Whereas loudness depends on the energy of the wave. ... The pitch of a sound depends on the frequency while loudness of a sound depends on the amplitude of sound waves.
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3 years ago
What 2 things need to be known about an object in order to determine its kinetic energy?
NikAS [45]
I believe the answer is the mass of the object and the speed at which it is moving. 
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3 years ago
What do solar cells convert into electricity?<br> wind<br> SERE<br> water<br> sunlight<br> plants
STatiana [176]
Hi am not sure but have searched on online
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At its Ames Research Center, NASA uses its large "20-G" centrifuge to test the effects of very large accelerations ("hypergravit
Over [174]

Answer:

v=32.9m/s

Explanation:

The acceleration needed to mantain a circular motion of radius r and speed v is given by the equation a=v^2/r

This is the centripetal acceleration. The person will feel what is called a centrifugal acceleration, of the same value, because he is not in an inertial frame (thus subject to fictitious forces, product of inertia).

We want to know the speed of his head when it is subject to 12.5 times the value of the acceleration of gravity while moving on a 8.84m radius circle, so we must do:

v=\sqrt{ar} = \sqrt{12.5gr}=\sqrt{(12.5)(9.8m/s)(8.84m)}=32.9m/s

7 0
3 years ago
A sound source is moving at 80 m/s toward a stationary listener that is standing in still air (a) Find the wavelength of the sou
Setler [38]

Answer:

a. wavelength of the sound, \vartheta = 1.315\vartheta_{o}

b. observed frequecy, \lambda = 0.7604\lambda_{o}

Given:

speed of sound source, v_{s} = 80 m/s

speed of sound in air or vacuum, v_{a} = 343 m/s

speed of sound observed, v_{o} = 0 m/s

Solution:

From the relation:

v = \vartheta \lambda        (1)

where

v = velocity of sound

\vartheta = observed frequency of sound

\lambda = wavelength

(a) The wavelength of the sound between source and the listener is given by:

\lambda = \frac{v_{a}}{\vartheta }         (2)

(b) The observed frequency is given by:

\vartheta = \frac{v_{a}}{v_{a} - v_{s}}\vartheta_{o}

\vartheta = \frac{334}{334 - 80}\vartheta_{o}

\vartheta = 1.315\vartheta_{o}                (3)

Using eqn (2) and (3):

\lambda = \frac{334}{1.315} = \frac{1}{1.315}\frac{v_{a}}{\vartheta_{o}}

\lambda = 0.7604\lambda_{o}

4 0
3 years ago
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