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exis [7]
3 years ago
6

A transverse wave is moving toward the right in a uniform medium. Point X represents a particle of the uniform medium. Which dia

gram represents the direction of the motion of particle X at the instant shown?

Physics
2 answers:
trapecia [35]3 years ago
6 0
After looking at the transverse waves in the diagram you listed above, the one diagram that does represent the direction of particle X at the instant show in diagram number 3. The direction of the wave motion is up. The correct answer choice will be 3. 
vladimir2022 [97]3 years ago
6 0
<h3><u>Answer</u>;</h3>

(4) last diagram

<h3><u>Explanation</u>;</h3>
  • <em><u>A transverse wave is a type of wave where the movement of the particles of the medium is perpendicular to the direction of the propagation of the wave. </u></em>
  • Transverse waves forms regions of maximum displacement known as crests and troughs. The crests are the top points of the wave and the trough is the bottom point of the wave.
  • In a transverse wave particles of the medium vibrate to and from in a direction perpendicular to the direction of energy transport.
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Helga [31]

Answer:

A: The acceleration is 7.7 m/s up the inclined plane.

B: It will take the block 0.36 seconds to move 0.5 meters up along the inclined plane

Explanation:

Let us work with variables and set

m=5kg\\\\F_H=100N\\\\\mu=0.3\\\\\theta=37^o.

As shown in the attached free body diagram, we choose our coordinates such that the x-axis is parallel to the inclined plane and the y-axis is perpendicular. We do this because it greatly simplifies our calculations.

Part A:

From the free body diagram we see that the total force along the x-axis is:

F_{tot}=mg*sin(\theta)+F_s-F_Hcos(\theta).

Now the force of friction is F_s=\mu*N, where N is the normal force and from the diagram it is F_y=mg*cos(\theta).

Thus F_s=\mu*N=\mu*mg*cos(\theta).

Therefore,

F_{tot}=mg*sin(\theta)+\mu*mg*cos(\theta)-F_Hcos(\theta)\\\\=mg(sin(\theta)+\mu*cos(\theta))-F_Hcos(\theta).

Substituting the value for F_H,m,\mu, and \:\theta we get:

F_{tot}= -38.63N.

Now acceleration is simply

a=\frac{F_H}{m} =\frac{-38.63N}{5kg} =-7.7m/s.

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Part B:

d=\frac{1}{2} at^2

Which can be rearranged to solve for t:

t=\sqrt{\frac{2d}{a} }

Substitute the value of d=0.50m and a=7.7m/s and we get:

t=0.36s.

which is our answer.

Notice that in using the formula to calculate time we used the positive value of a, because for this formula absolute value is needed.

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4 years ago
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