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3 years ago

11

One string of a certain musical instrument is 75.0cm long and has a mass of 8.77g . It is being played in a room where the speed

of sound is 344 m/s.
To what tension must you adjust the string so that, when vibrating in its second overtone, it produces sound of wavelength 0.761m ?
(Assume that the breaking stress of the wire is very large and isn’t exceeded.)

1 answer:

zheka24 [161]3 years ago

3 0

The question is asking to calculate the tension that the string has to adjust the string so that when vibrating in its second overtone, it produces sound of wavelength of 0.761m, base on my calculation, the calculation must be done by the formula of <span>v=λf</span><span>., I hope this would help </span>

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2 years ago

A car initially traveling at 24 m/s slams on the brakes and moves forward 196 m before coming to a complete halt. What was the m

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Answer:

$-1.47 m/s^2$

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2 years ago

child slides down a snow‑covered slope on a sled. At the top of the slope, her mother gives her a push to start her off with a s

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Answer:

θ = 13.7º

Explanation:

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ΔU, is the change in the gravitational potential energy.
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$\Delta U = 0 - m*g*h = -m*g*d* sin\theta (3)$

Finally, the work done for non-conservative forces, is the work done by the friction force, along the slope, as follows:

$W_{nc} = F_{f} * d * cos 180\º \\\\ = 0.2*m*g*d* cos 180\º = -0.2*m*g*d (4)$

Replacing (2), (3), and (4) in (1), simplifying common terms, and rearranging, we have:

$\frac{1}{2}* (v_{f} ^{2} - v_{0} ^{2}) = g*d* sin\theta -0.2*g*d$

Replacing by the givens and the knowns, we can solve for sin θ, as follows: $\frac{1}{2}*( (4.30 m/s) ^{2} - (0.75 m/s)^{2}) = 9.8 m/s2*25.5m* sin\theta -0.2*9.8m/s2*25.5m\\ \\ 8.56 (m/s)2 = 250(m/s)2* sin \theta -50 (m/s)2\\ \\ sin \theta = \frac{58.6 (m/s)2}{250 (m/s)2} = 0.236$ ⇒ θ = sin⁻¹ (0.236) = 13.7º

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2 years ago

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Answer:

Part a)

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Part B)

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Part c)

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Explanation:

Part a)

As we know that

$a = R \alpha$

so we will have

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$\alpha = \frac{a}{R}$

$\alpha = \frac{1.80}{0.230 \times 10^{-2}}$

$\alpha = 782.6 rad/s^2$

Part B)

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$\omega = \alpha t$

so we have

$\omega = 782.6 \times 0.750$

$\omega = 587 rad/s$

Part c)

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