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aliina [53]
3 years ago
12

Two violinists are trying to play in tune. However, whenever they play their A string at the same time they hear a beat frequenc

y of 5.0 Hz. One of the violinists measures with an electronic tuner and finds her A string is at 462 Hz. What are the possible frequencies for the A string of the other violinist?(multiple answers)
462 Hz
452 Hz
457 Hz
467 Hz
472 Hz
Physics
1 answer:
kaheart [24]3 years ago
4 0

Answer:

The possible frequencies for the A string of the other violinist is 457 Hz and 467 Hz.

(3) and (4) is correct option.

Explanation:

Given that,

Beat frequency f = 5.0 Hz

Frequency f'= 462 Hz

We need to calculate the possible frequencies for the A string of the other violinist

Using formula of frequency

f'=f_{1}-f...(I)

f'=f_{1}+f...(II)

Where, f= beat frequency

f₁ = frequency

Put the value in both equations

f'=462-5=457\ Hz

f'=462+5=467\ Hz

Hence, The possible frequencies for the A string of the other violinist is 467 Hz and 457 Hz.

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now we have

E = \frac{4kq b}{(b^2 + \frac{d^2}{2})^{1.5}}

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3 years ago
What are the three longest wavelengths for standing waves on a 264- cm -long string that is fixed at both ends
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The three longest wavelengths for the standing waves on a 264-cm long string that is fixed at both ends are:

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Given data:

Length of the fixed string = 264cms = 2.64 meters

The wavelength for standing waves is given by:

λ = 2L/n

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For n = 1,

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For n = 2,

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For n = 3,

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= 1.7 meters

To learn more about standing waves: brainly.com/question/14151246

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In an elastic head-on collision, a 0.60 kg cart moving at 5.0 m/s [W] collides with a 0.80 kg cart moving at 2.0 m/s [E]. The co
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Answer:

The answer is given below

Explanation:

u is the initial velocity, v is the final velocity. Given that:

m_1=0.6kg,u_1=-5m/s(moving \ west),m_2=0.8kg,u_2=2m/s,k=1200N/m

a)

The final velocity of cart 1 after collision is given as:

v_1=(\frac{m_1-m_2}{m_1+m_2})u_1+\frac{2m_2}{m_1+m_2}u_2\\  Substituting:\\v_1=\frac{0.6-0.8}{0.6+0.8} (-5)+\frac{2*0.8}{0.6+0.8}(2)= 5/7+16/7=3\ m/s

The final velocity of cart 2 after collision is given as:

v_2=(\frac{m_2-m_1}{m_1+m_2})u_2+\frac{2m_1}{m_1+m_2}u_1\\  Substituting:\\v_1=\frac{0.8-0.6}{0.6+0.8} (2)+\frac{2*0.6}{0.6+0.8}(-5)= 2/7-30/7=-4\ m/s

b) Using the law of conservation of energy:

\frac{1}{2}m_1u_1+ \frac{1}{2}m_2u_2=\frac{1}{2}m_1v_1+\frac{1}{2}m_2v_2+\frac{1}{2}kx^2\\x=\sqrt{\frac{m_1u_1+m_2u_2-m_1v_1-m_2v_2}{k}}\\ Substituting\ gives:\\x=\sqrt{\frac{0.6*(-5)^2+0.8*2^2-(0.6*3^2)-(0.8*(-4)^2)}{1200}}=\sqrt{0}=0\ cm

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