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3 years ago

Q) Suppose, you are in a sporting event. You notice that everyone stands up when it’s his turn,

Physics

1 answer:

zysi [14]3 years ago

3 0

Answer:

The nature of the wave formed is a transverse progressive wave.

Explanation:

A wave is a disturbance that travels through a material medium without permanent displacement of the particles of the medium. The two major types are: transverse and longitudinal.

A transverse wave is one in which the direction of vibration of the particles of the medium is perpendicular to the direction of propagation of the wave. Examples are: water wave, light wave etc. While a longitudinal wave is one in which the direction of vibrations of the particles of the medium is parallel with the direction of propagation of the wave, creating a region of rarefaction and compression. Examples are; sound wave, wave in a rope, wave in a slinky etc.

The cited wave formed in the given question is a transverse wave because each person stands and sits after some time to create a moving (progressive) wave without them moving from their positions.

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You are climbing in the high sierra when you suddenly find yourself at the edge of a fog-shrouded cliff. to find the height of t

sergey [27]

The total time it takes the sound to reach our on top of the cliff (10.0 s) is actually sum of two times:
- the time it takes the rock to hit the ground starting from the top of the cliff, t1
- the time it takes the sound to reach the top of the cliff starting from the ground, t2
1) The rock moves by uniformly accelerated motion, with constant acceleration $g=9.81 m/s^2$ . Its law of motion is given by $y(t)=h-v_0t-\frac{1}{2}gt^2$ , where $v_0=0$ is the initial velocity of the rock, and h is the height of the cliff. The time t1 is the time at which the rock reaches the ground, so that y(t1)=0, and the equation becomes
$0=h-\frac{1}{2}gt_1^2$
2) The sound moves from the ground to the top of the cliff by uniform motion, with constant speed v=343 m/s. Therefore, the sound covers the distance h (the height of the cliff) in a time t2 given by
$h=vt_2$
3) If we rewrite h in both equations, we can write:
$=\frac{1}{2}gt_1^2=vt_2$ (1)
4) We also know that the sum of t1 and t2 is equal to 10 seconds:
$t_1+t_2=10$
from which we find
$t_2=10-t_1$
if we substitute this into eq.(1), we get
$\frac{1}{2}gt_1^2+vt_1-10v=0$
Numerically:
$4.9t_1^2 + 343 t_1 - 3430=0$
Solving the equation, we find the solution $t_1=8.87 s$ (the other solution is negative, so it does not have physical meaning). As a consequence,
$t_2 = 10-t_1 = 1.13 s$
and the height of the cliff is given by
 $h=vt_2=(343 m/s)(1.13 s)=388 m$

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SCORPION-xisa [38]

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Evgesh-ka [11]

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Luda [366]

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PIT_PIT [208]

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