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Oksi-84 [34.3K]

3 years ago

12

Lydia is 9kg underweight. Caroline has an ideal weight for her height. Viv is 8kg overweight. Who is most likely to have the hig

hest blood pressure?

2 answers:

Crank3 years ago

5 0

Answer: Viv

Explanation:

UkoKoshka [18]3 years ago

4 0

Viv most likely —————-

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Two loudspeakers are located 4.965 m apart on an outdoor stage. A listener is 12.608 m from one and 18.368 m from the other. Dur

leonid [27]

Answer:

Approximately $30.0\; \rm Hz$ .

Explanation:

Look up the speed of sounds in the air at $25\; ^\circ\rm C$ : $v \approx 346\; \rm m \cdot s^{-1}$ .

Let the frequency of this tune be $f\; \rm Hz$ . The wavelength of the tune would be $\displaystyle \lambda = \frac{v}{f} \approx \frac{346}{f}$ .

The distance between the first speaker and the listener is $12.608\; \rm m$ . How many wavelengths can fit into that distance?

$\displaystyle \frac{12.608}{\lambda} \approx \frac{12.608}{346 / f} = \frac{12.608}{346} \cdot f$ .

Similarly, the distance between the second speaker and the listener is $18.368\; \rm m$ . Number of wavelengths in that distance:

$\displaystyle \frac{18.368}{\lambda} \approx \frac{18.368}{346 / f} = \frac{18.368}{346} \cdot f$ .

Difference between these two numbers:

$\begin{aligned} &\frac{18.368}{346} \cdot f - \frac{12.608}{346} \cdot f = \frac{5.76}{346}\cdot f\end{aligned}$ .

For destructive interference to occur, that difference should be equal to $\displaystyle \frac{1}{2}$ , $\displaystyle 1 + \frac{1}{2}= \frac{3}{2}$ , $\cdots$ , or $\displaystyle \left(k+ \frac{1}{2}\right)$ in general ( $k$ can be any non-negative whole number.)

Let $\begin{aligned} \frac{5.76}{346}\cdot f = k + \frac{1}{2}\end{aligned}$ , and solve for $f$ .

$\begin{aligned} f&= \left.\left(k + \frac{1}{2}\right) \right/\frac{5.76}{346}\\ &= \frac{346}{5.76} \, k + \frac{1}{2} \times \frac{346}{5.76} \\ &\approx 60.1\, k + 30.0 \end{aligned}$ .

The next step is to find the values of $k$ that ensure $20 \le f \le 20\times 10^{3}$ (frequency is between $20\; \rm Hz$ and $20\, \rm kHz$ .) It turns out that $k = 0$ (the smallest $k$ value possible) would be sufficient. In that case, the frequency is approximately $30.0\; \rm Hz$ . Using a larger $k$ would only increase the frequency.

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The speed of a sound wave depends mostly on ?

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<span>The medium in which it travels through</span>

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