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klio [65]
3 years ago
9

A violin string is 45.0 cm long and has a mass of 0.242 g. When tightened on the neck of the violin, the distance between the pi

n and fret is 42.5 cm. If the string is tightened to play a 'Concert A' of 440. Hz as its fundamental/ first harmonic, what is the tension in the violin string
Physics
1 answer:
stiks02 [169]3 years ago
3 0

Answer:

The tension is 75.22 Newtons

Explanation:

The velocity of a wave on a rope is:

v=\sqrt{\frac{TL}{M}} (1)

With T the tension, L the length of the string and M its mass.

Another more general expression for the velocity of a wave is the product of the wavelength (λ) and the frequency (f) of the wave:

v= \lambda f (2)

We can equate expression (1) and (2):

\sqrt{\frac{TL}{M}}=\lambda f

Solving for T

T= \frac{M(\lambda f)^2}{L} (3)

For this expression we already know M, f, and L. And indirectly we already know λ too. On a string fixed at its extremes we have standing waves ant the equation of the wavelength in function the number of the harmonic N_{harmonic} is:

\lambda_{harmonic}=\frac{2l}{N_{harmonic}}

It's is important to note that in our case L the length of the string is different from l the distance between the pin and fret to produce a Concert A, so for the first harmonic:

\lambda_{1}=\frac{2(0.425m)}{1}=0.85 m

We can now find T on (3) using all the values we have:

T= \frac{2.42\times10^{-3}(0.85* 440)^2}{0.45}

T=75.22 N

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xenn [34]

Answer:

From certain assumptions that the walking speed is 2 m/s, and the stop time is 0.1 s the acceleration would be -20 m/s

Explanation:

Using the average acceleration formula:

a=\frac{\Delta v}{\Delta t} where \Delta v and \Delta t are the changes in the speed and time respectively.

We have by assuming that the walking speed is 2 m/s and the stop time is 0.1s which is equal to the change in time during the stopping.

\Delta v=v_f-v_i=0-2 m/s=-2 m/s, where v_i,v_f are the initial speed and final speed respectively, and \Delta t=0.1 s

Plugging the previous in the average acceleration formula we get

a=\frac{-2}{0.1}=-20\, m/s where the minus sign indicates an acceleration in the opposite direction of the motion (or in other word opposite to the speed's direction).

4 0
3 years ago
What are the formulas for Work, KE, GPE and Power?
Katyanochek1 [597]

Answer:

work=f×d

GPE =m×g×h

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6 0
3 years ago
A sound having a frequency of 395 Hz travels through air at 331 m/s. What is the wavelength of the sound? Answer in units of m.
jenyasd209 [6]
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8 0
3 years ago
2. Which of the following is an example of work being done on an object? A prism scatters ultraviolet light into visible light.
Liula [17]
A man pushes a couch across the room is the answer!

7 0
2 years ago
Read 2 more answers
What work is done by the electric force when the charge moves a distance of 2.70 m at an angle of 45.0∘ downward from the horizo
Andreyy89

Incomplete question as many data is missing.I have assumed value of charge and electric field.The complete question is here

A charge of 28 nC is placed in a uniform electric field that is directed vertically upward and that has a magnitude of 5.00×10⁴ V/m.

What work is done by the electric force when the charge moves a distance of 2.70 m at an angle of 45.0 degrees  downward from the horizontal?

Answer:

W_{work}=2.67*10^{-3}J

Explanation:

Given data

Charge q=28 nC

Electric field E=5.00×10⁴ V/m.

Distance d=2.70 m

Angle α=45°

To find

Work done by electric force

Solution

W_{work}=F_{force}*D_{distance}Cos\alpha  \\where\\F_{force}=q_{charge}*E_{Electric-Field}\\So\\W_{work}=qE*D*Cos\alpha \\W_{work}=(28*10^{-9}C )(5.00*10^{4}V/m )(2.70m)Cos(45)\\W_{work}=2.67*10^{-3}J

8 0
3 years ago
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