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1 year ago

10

Which diagram shows the resulting wave when they cross at point X?

1 answer:

zvonat [6]1 year ago

5 0

The resulting wave will have zero amplitude, thus it will form a straight line.

<h3>What is a crest?</h3>

A crest is an upward displacement of a progressive wave.

<h3>What is a trough?</h3>

A trough is a downward displacement of a progressive wave.

<h3>Superimposition of the two waves</h3>

When the two waves meet each other at point X, they will superimpose and cancel out since they have equal amplitude and are in opposite direction.

The resulting wave will have zero amplitude, thus it will form a straight line.

Learn more about crest here: brainly.com/question/1430231

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3 years ago

A 1000 kg elevator is rising and its speed is increasing with an acceleration of 3 m/s^2. What is the resulting tension in the v

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2 years ago

A particle is moving with SHM of period pie . initially it is 10 cm from The center of the motion and moving in the positive dir

Viefleur [7K]

Answer:

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y = Acos(2πt/T + φ) = Acos(2πt/π + φ) = Acos(2t + φ)

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3 years ago

A trick shot archer shoots an arrow with a velocity of 30.0 m/s at an angle of 20.0 degrees with respect to the horizontal. An a

svlad2 [7]

Answer:

$u'=10.259\ m.s^{-1}$ is the initial velocity of tossing the apple.

the apple should be tossed after $\Delta t=0.0173\ s$

Explanation:

Given:

velocity of arrow in projectile, $v=30\ m.s^{-1}$

angle of projectile from the horizontal, $\theta=20^{\circ}$

distance of the point of tossing up of an apple, $d=30\ m$

<u>Now the horizontal component of velocity:</u>

$v_x=v\ cos\ \theta$

$v_x=30\times cos\ 20^{\circ}$

$v_x=28.191\ m.s^{-1}$

<u>The vertical component of the velocity:</u>

$v_y=v.sin\ \theta$

$v_y=30\times sin\ 20^{\circ}$

$v_y=10.261\ m.s^{-1}$

<u>Time taken by the projectile to travel the distance of 30 m:</u>

$t=\frac{d}{v_x}$

$t=\frac{30}{28.191}$

$t=1.0642\ s$

<u>Vertical position of the projectile at this time:</u>

$h=v_y.t-\frac{1}{2}g.t^2$

$h=10.261\times 1.0642-\frac{1}{2} \times 9.8\times 1.0642^2$

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$v'^2=u'^2-2g.h$

$0^2=u'^2-2\times 9.8\times 5.3701$

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<u>Time taken to reach this height:</u>

$v'=u'-g.t'$

$0=10.259-9.8\times t'$

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<u>We observe that </u> $t>t'$ <u> hence the time after the launch of the projectile after which the apple should be tossed is:</u>

$\Delta t=t-t'$

$\Delta t=1.0642-1.0469$

$\Delta t=0.0173\ s$

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3 years ago

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Explanation:

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3 years ago