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Murrr4er [49]
3 years ago
15

If the strings have the same thickness butdifferent lengths, which of the following parameters, if any, willbe different in the

two strings?Check all that apply.wave frequencywave speedwavelengthnone of the above
Physics
1 answer:
gizmo_the_mogwai [7]3 years ago
7 0

Answer:

1.Length is one of the four factors on which the wave frequency depends. So if the length of the string changes then there will be a change in the vibration of string. So in this case if the lengths are different then the wave frequency of both will be different.

2. Wave speed will be the same as it depends on tension and linear density of it.

3.  Wavelength itself is find out by the length of string so it depends on length and it will vary with the lengths of strings.

Explanation:

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a) v_{1}=\frac{(62.5-66)ft}{(2.5-2)s}=-7ft/s

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y=65t-16t^{2}

a) To find the average velocity at the given time we need to use the following formula:

v=\frac{y_{2}-y_{1}  }{t_{2}-t_{1}  }

Being said that, we need to find the ball's position at t=2, t=2.5, t=2.1, t=2.01, t=2.001

y_{t=2}=65(2)-16(2)^{2} =66ft

y_{t=2.5}=65(2.5)-16(2.5)^{2} =62.5ft

v_{1}=\frac{(62.5-66)ft}{(2.5-2)s}=-7ft/s

--

y_{t=2.1}=65(2.1)-16(2.1)^{2} =65.94ft

v_{2}=\frac{(65.94-66)ft}{(2.1-2)s}=-0.6ft/s

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y_{t=2.01}=65(2.01)-16(2.01)^{2} =66.0084ft

v_{3}=\frac{(66.0084-66)ft}{(2.01-2)s}=0.84ft/s

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y_{t=2.001}=65(2.001)-16(2.001)^{2} =66.001ft

v_{4}=\frac{(66.001-66)ft}{(2.001-2)s}=1ft/s

b) To find the instantaneous velocity we need to derivate the equation

v=\frac{df}{dt}=65-32t

v=65-32(2)=1ft/s

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