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Aleks04 [339]
3 years ago
15

As a wave moves through a medium, particles are displaced and (2 points)

Physics
1 answer:
Scorpion4ik [409]3 years ago
7 0

As a wave moves through a medium, particles are displaced and return to their normal position after the wave passes.

Explanation:

A wave is a traveling disturbance that  carries energy from one location to  another. All waves move in straight lines  outward and away from the source of a  disturbance. Like the radiating circular  ripples, the waves of water carry energy  away from where a rock was dropped into  the pond.

Waves can move as a single pulse or as a continuous series of waves, carrying  energy away from its source. A pulse is a single disturbance, wave, or ripple that moves  outward from the point of disturbance. A train of waves are many waves emitted over and  over again from a single source.

As waves travel through matter, they will temporarily  displace the molecules or particles in matter up-and-down  or side-to-side. Waves move the energy but they do not  carry the matter with them longitudinally as they move  through matter. Once the disturbance passes, the medium  will return to its original state or position.

Therefore, as the waves move through a medium, particles are displaced and return to their normal position after the wave passes.

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the maximum range of a projectile is 2÷√3 times its actual range what is the angle of the projection for the actual range​
Murrr4er [49]

Answer:

The actual angle is 30°

Explanation:

<h2>Equation of projectile:</h2><h2>y axis:</h2>

v_y(t)=vo*sin(A)-g*t

the velocity is Zero when the projectile reach in the maximum altitude:

0=vo-gt\\t=\frac{vo}{g}

When the time is vo/g the projectile are in the middle of the range.

<h2>x axis:</h2>

d_x(t)=vo*cos(A)*t\\

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R=d_x(t=2*\frac{vo}{g})

R=vo*cos(A)*2\frac{vo}{g} \\\\R=\frac{(vo)^{2}*2* sin(A)cos(A)}{g} \\\\R=\frac{(vo)^{2} sin(2A)}{g}

**sin(2A)=2sin(A)cos(A)

<h2>The maximum range occurs when A=45°(because sin(90°)=1)</h2><h2>The actual range R'=(2/√3)R:</h2>

Let B the actual angle of projectile

\frac{vo^{2} }{g} =(\frac{2}{\sqrt{3} }) \frac{vo^{2} *sin(2B)}{g}\\\\1= \frac{2 }{\sqrt{3}} *sin(2B)\\\\sin(2B)=\frac{\sqrt{3}}{2}\\\\

2B=60°

B=30°

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