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sesenic [268]
3 years ago
14

The speed of sound at room temperature (20 degrees Celsius) is 343 m/s. If the speed of sound in air increases about 0.60 m/s fo

r every 1 degrees Celsius increase, what is the temperature when the speed of sound is 353 m/s?
Physics
1 answer:
stepan [7]3 years ago
4 0

Answer:

36,67 degrees Celsius

Explanation:

The simplest way to approach this problem, given the information provided, is to simply start with the speed difference.

Goal: 353 m/s

Start: 343 m/s (at 20 degrees Celsius).

Difference: 10 m/s

Variation rate: 0.60 m/s/d (d = degree)

d = \frac{10 m/s}{0.60 m/s/d}  = 16,67 d

So, 16,67 degrees more than the starting point.

The temperature will then be 36.67 degrees Celsius, when the sound travels at the speed of 353 m/s.

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Recall that average velocity is equal to change in position over a given time interval,

\vec v_{\rm ave} = \dfrac{\Delta \vec r}{\Delta t}

so that the <em>x</em>-component of \vec v_{\rm ave} is

\dfrac{x_2 - (-2.25\,\mathrm m)}{1.60\,\mathrm s} = 2.70\dfrac{\rm m}{\rm s}

and its <em>y</em>-component is

\dfrac{y_2 - 5.70\,\mathrm m}{1.60\,\mathrm s} = -2.50\dfrac{\rm m}{\rm s}

Solve for x_2 and y_2, which are the <em>x</em>- and <em>y</em>-components of the copter's position vector after <em>t</em> = 1.60 s.

x_2 = -2.25\,\mathrm m + \left(2.70\dfrac{\rm m}{\rm s}\right)(1.60\,\mathrm s) \implies \boxed{x_2 = 2.07\,\mathrm m}

y_2 = 5.70\,\mathrm m + \left(-2.50\dfrac{\rm m}{\rm s}\right)(1.60\,\mathrm s) \implies \boxed{y_2 = 1.70\,\mathrm m}

Note that I'm reading the given details as

x_1 = -2.25\,\mathrm m \\\\ y_1 = -5.70\,\mathrm m \\\\ v_x = 2.70\dfrac{\rm m}{\rm s}\\\\ v_y=-2.50\dfrac{\rm m}{\rm s}

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Answer:

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I have corrected this question and now

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