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Ivanshal [37]
3 years ago
9

It is found that a 5.70 m segment of a long string contains three complete waves and has a mass of 180 g. The string is vibratin

g sinusoidally with a frequency of 55.0 Hz and a peak-to-valley distance of 19.0 cm. (The "peak-to-valley" distance is the vertical distance from the farthest positive position to the farthest negative position). Calculate the wavelenght.
Physics
2 answers:
iragen [17]3 years ago
8 0

Answer:

wavelength = 3.8 m

Explanation:

As we know that linear mass density is defined as the ratio of mass and length

so here we have

\mu = \frac{m}{L}

\mu = \frac{0.180}{5.70}

now we have

\mu = 0.0315 kg/m

Now it is given that string contains three complete waves

length of one segment on string is half of the wavelength

so here we have

3\frac{\lambda}{2} = 5.70 m

\lambda = 3.8 m

So wavelength of the wave on string is 3.8 m

Hoochie [10]3 years ago
3 0

Answer:

1.9 m.

Explanation:

Three complete waves in the length of 5.7 m

The distance traveled by one complete wave is called wavelength.

Thus, the distance traveled by one wave = 5.7 / 3 = 1.9 m

Thus, the wavelength is 1.9 m.

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lisabon 2012 [21]

Magnitude of acceleration = (change in speed) / (time for the change) .

Change in speed = (ending speed) - (starting speed)

                            =       zero            - (43 m/s)

                            =          -43 m/s .

Magnitude of acceleration = (-43 m/sec) / (0.28 sec)

                                          =  (-43 / 0.28)  (m/sec) / sec

                                          =        153.57...  m/s²

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5 0
3 years ago
Read 2 more answers
1. I drop a penny from the top of the tower at the front of Fort Collins High School and it takes 1.85 seconds to hit the ground
Svetradugi [14.3K]
You have three known variables:

Acceleration - 9.8m/s^2
Time - 1.85s
Initial Velocity - 0m/s

For the first part of your question:

v = u + at

v = (0)+(9.8)(1.1)

v = 10.78 m/s  v=11 m/s (2 significant figures)

For the second part of your question:

v=u+at

v=(0)+(9.8)(1.85)
v=18.13 m/s



This still needs to be converted to m/h:

18.13m/s = 18.13\times 3600 metres/h=65628  metres/hour

65628 metres/hour = 65628\div1600 mi/h = 40.7925 mi/h

= 41 mi/hr (2 significant figures)





5 0
3 years ago
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What is the displacement of the GBS cross-country team if they begin at the school, run 10 miles, and finish back at the school?
olga2289 [7]

Answer:

0

Explanation:

8 0
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Using a density of air to be 1.21kg/m3, the diameter of the bottom part of the filter as 0.15m (assume circular cross-section),
salantis [7]

Answer:

The  drag coefficient is  D_z  =  1.30512  

Explanation:

From the question we are told that

     The density of air is  \rho_a  = 1.21 \ kg/m^3

     The diameter of bottom part is  d = 0.15 \ m

The  power trend-line  equation is mathematically represented as

      F_{\alpha }  = 0.9226 * v^{0.5737}

let assume that the velocity is  20 m/s

Then

      F_{\alpha }  = 0.9226 * 20^{0.5737}

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The drag coefficient is mathematically represented as

      D_z  =  \frac{2 F_{\alpha } }{A \rho v^2 }

Where  

     F_{\alpha } is the drag force

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substituting values

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Then

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   D_z  =  1.30512  

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We know V=IR (Ohm's law).

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V=18V
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