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morpeh [17]
3 years ago
14

A series of pulses of amplitude 0.28 m are sent down a string that is attached to a post at one end. The pulses are reflected at

the post and travel back along the string without loss of amplitude. What is the amplitude at a point on the string where two pulses cross if the string is rigidly attached to the post?
Physics
2 answers:
andreyandreev [35.5K]3 years ago
5 0

Answer:

The answer is zero displacement (0 m)

Explanation:

If the end of the string is not free, the reflected pulse has the same amplitude but opposite polarity to the incident pulse. Because of this, for A, the result equals zero (0 m).

irina1246 [14]3 years ago
3 0

Answer:

a. The pulses cancel each other resulting in zero displacement.

b. The pulses reinforce each other, having a displacement of 2× amplitude or 0.56 m.

Explanation:

When the pulse are sent down the attached string  it gets reflected and we have crossing pulses as the incident and reflected pulses cross each other.

a. If the string is rigidly attached to the post then the incident and reflected pulses will have the same amplitude but different direction. That is either the reflected will be going up and the incident down thereby resulting in a cancellation or zero displacement

Incident amplitude = 0.28 m

Reflected amplitude = -0.28 m

Sum = 0.28 m - 0.28 m = 0 m

b. If the end at which the relfection occurs is free to slide, then the incident and the reflected pulses will again have the same amplitude and in this case, the same direction. Therefore;

Incident amplitude = 0.28 m

Reflected amplitude = 0.28 m

Sum = 0.28 m + 0.28 m = 0.56 m

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SOVA2 [1]

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because of that she will experience a varying electrical field, and thus a varying electric current will be zero.

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Hence option C is correct.

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2 years ago
A uniform disk with a mass of 5.0 kg and diameter 30 cm rotates on a frictionless fixed axis through its center and perpendicula
igomit [66]

Answer:

Angular acceleration of the disk will be \alpha =10.714rad/sec^2

Explanation:

We have given mass of the disk m = 5 kg

Diameter of the disk d = 30 cm = 0.3 m

So radius r=\frac{d}{2}=\frac{0.3}{2}=0.15m

Moment of inertia of disk is given by I=\frac{1}{2}mr^2=\frac{1}{2}\times 5\times 0.15^2=0.056kgm^2

Force is given by F=4 N

Torque is given as \tau =Fr=4\times 0.15=0.6N-m

We also know that torque is given by \tau =I\alpha

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A 4.80 −kg ball is dropped from a height of 15.0 m above one end of a uniform bar that pivots at its center. The bar has mass 7.
Margarita [4]

Answer:

h = 13.3 m

Explanation:

Given:-

- The mass of ball, mb = 4.80 kg

- The mass of bar, ml = 7.0 kg

- The height from which ball dropped, H = 15.0 m

- The length of bar, L = 6.0 m

- The mass at other end of bar, mo = 5.10 kg

Find:-

The dropped ball sticks to the bar after the collision.How high will the other ball go after the collision?

Solution:-

- Consider the three masses ( 2 balls and bar ) as a system. There are no extra unbalanced forces acting on this system. We can isolate the system and apply the principle of conservation of angular momentum. The axis at the center of the bar:

- The angular momentum for ball dropped before collision ( M1 ):

                                 M1 = mb*vb*(L/2)

Where, vb is the speed of the ball on impact:

- The speed of the ball at the point of collision can be determined by using the principle of conservation of energy:

                                  ΔP.E = ΔK.E

                                  mb*g*H = 0.5*mb*vb^2

                                  vb = √2*g*H

                                  vb = ( 2*9.81*15 ) ^0.5

                                  vb = 17.15517 m/s

- The angular momentum of system before collision is:

                                  M1 = ( 4.80 ) * ( 17.15517 ) * ( 6/2)

                                  M1 = 247.034448 kgm^2 /s

- After collision, the momentum is transferred to the other ball. The momentum after collision is:

                                  M2 = mo*vo*(L/2)

- From principle of conservation of angular momentum the initial and final angular momentum remains the same.

                                 M1 = M2

                                 vo = 247.03448 / (5.10*3)

                                 vo = 16.14604 m/s

- The speed of the other ball after collision is (vo), the maximum height can be determined by using the principle of conservation of energy:

                                  ΔP.E = ΔK.E

                                  mo*g*h = 0.5*mo*vo^2

                                  h = vo^2 / 2*g

                                  h = 16.14604^2 / 2*(9.81)

                                  h = 13.3 m

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