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andrezito [222]
3 years ago
5

N what direction does the medium move relative to the direction of the wave? Explain.

Physics
2 answers:
Ganezh [65]3 years ago
6 0
At the point when a ultrasonic pulse, containing, say, ten semi sinusoidal motions, is reflected in air from a harsh surface, it is watched tentatively that the scattered wave prepare contains disengagements, which are firmly similar to those found in blemished precious stones. We demonstrate hypothetically that such disengagements are not out of the ordinary at whatever point restricted trains of waves, eventually got from a similar oscillator, go in various headings and meddle - for instance in a disseminating issue. Scattering is not included. Conditions are given demonstrating the itemized structure of edge, screw and blended edge-screw separations, and furthermore of parallel arrangements of such disengagements. Edge disengagements can float with respect to the wave prepare at any speed; they can likewise climb, and screw separations can coast. Wavefront separations might be bended, and they may meet; they may impact and bounce back; they may destroy each other or be made as circles or combines. With separations in wave trains, not at all like precious stone disengagements, there is no breakdown of linearity close to the inside. Scientifically they are lines along which the stage is vague; this suggests the wave abundancy is zero.(for a wide range of waves)

I hope I helped!
Mashcka [7]3 years ago
3 0
Two types of mechanical waves: longitudinal<span> waves and </span>transverse<span> waves; the medium movement differs between the two.

</span>In a longitudinal wave the medium particle movement is parallel to the direction of wave propagation; example is sound wave in air.

I<span>n a transverse wave the medium particle movement is perpendicular to the direction of wave propagation; example is mechanical wave on a string.
</span><span>



</span>
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An advertisement for an all-terrain vehicle (ATV) claims that the ATV can climb inclined slopes of 35°. The minimum coefficient of static friction needed for this claim to be possible is 0.7

In an inclined plane, the coefficient of static friction is the angle at which an object slide over another.  

As the angle rises, the gravitational force component surpasses the static friction force, as such, the object begins to slide.

Using the Newton second law;

\sum F_x = \sum F_y = 0

\mathbf{mg sin \theta -f_s= N-mgcos \theta = 0 }

  • So; On the L.H.S

\mathbf{mg sin \theta =f_s}

\mathbf{mg sin \theta =\mu_s N}

  • On the R.H.S

N = mg cos θ

Equating both force component together, we have:

\mathbf{mg sin \theta =\mu_s \ mg \ cos \theta}

\mathbf{sin \theta =\mu_s \ \ cos \theta}

\mathbf{\mu_s = \dfrac{sin \theta }{ cos \theta}}

From trigonometry rule:

\mathbf{tan \theta= \dfrac{sin \theta }{ cos \theta}}

∴

\mathbf{\mu_s =\tan \theta}}

\mathbf{\mu_s =\tan 35^0}}

\mathbf{\mu_s = 0.700}}

Therefore, we can conclude that the minimum coefficient of static friction needed for this claim to be possible is 0.7

Learn more about static friction here:

brainly.com/question/24882156?referrer=searchResults

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