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Serggg [28]
3 years ago
10

Mac and Keena are experimenting with pulses on a rope. Mac vibrates one end up and down while Keena holds the other end. This cr

eates a pulse which they observe moving from end to end. How does the position of a point on the rope before the start of the pulse compare to its position after the pulse passes? Explain your reasoning.
Physics
1 answer:
Mariulka [41]3 years ago
4 0
Mac and Keena are experimenting with pulses on a rope. Mac vibrates one end up and down while Keena holds the other end. This creates a pulse which they observe moving from end to end. How does the position of a point on the rope before the start of the pulse compare to its position after the pulse passes? Explain your reasoning.
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DO NOT ANSWER IF YOU DON'T KNOW
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The last one, the soil will become weak & unable to support plant growth
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I would love to stretch a wire from our house to the Shop so I can 'call' my husband in for meals. The wire could be tightened t
dezoksy [38]
Note: I'm not sure what do you mean by "weight 0.05 kg/L". I assume it means the mass per unit of length, so it should be "0.05 kg/m".

Solution:
The fundamental frequency in a standing wave is given by
f= \frac{1}{2L} \sqrt{ \frac{T}{m/L} }
where L is the length of the string, T the tension and m its mass. If  we plug the data of the problem into the equation, we find
f= \frac{1}{2 \cdot 24 m} \sqrt{ \frac{240 N}{0.05 kg/m} }=1.44 Hz

The wavelength of the standing wave is instead twice the length of the string:
\lambda=2 L= 2 \cdot 24 m=48 m

So the speed of the wave is
v=\lambda f = (48 m)(1.44 Hz)=69.1 m/s

And the time the pulse takes to reach the shop is the distance covered divided by the speed:
t= \frac{L}{v}= \frac{24 m}{69.1 m/s}=0.35 s
7 0
3 years ago
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