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3 years ago

The distance from one compression in a sound wave to the next compression is defined as the __________.

Physics

1 answer:

a_sh-v [17]3 years ago

5 0

It is B) Wavelength, because it measures the "length" between "waves", therefore, wavelength.

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Find the period of the leg of a man who is 1.83 m in height with a mass of 67 kg. The moment of inertia of a cylinder rotating a

In-s [12.5K]

Answer:

$T = 1.108\,s$

Explanation:

The period of a physical pendulum is:

$T = \sqrt{\frac{I_{O}}{m\cdot g \cdot L} }$

$T=2\cdot \pi \sqrt{\frac{\frac{1}{3}\cdot m \cdot L^{2} }{m\cdot g\cdot L} }$

$T=2\cdot \pi \sqrt{\frac{L }{3\cdot g} }$

The length of the leg is approximately the height of the person:

$L = 0.915\,m$

The period is:

$T = 2\cdot \pi \sqrt{\frac{0.915\,m}{3\cdot (9.807\,\frac{m}{s^{2}} )} }$

$T = 1.108\,s$

4 0

3 years ago

Pulmonary circulation involves blood flow to and from the heart and the ____?

olga nikolaevna [1]

I believe the answer would be lungs

3 0

3 years ago

Read 2 more answers

Which is heavier, humid air or dry air ?

zvonat [6]

Answer:

dry air is way heaver

Explanation:

5 0

3 years ago

A light wave passes through an aperture (that is, a narrow slit). When it does so, the degree to which the wave spreads out will

crimeas [40]

Explanation:

Single slit diffraction

Diffraction is the phenomenon of spreading out of waves as they pass through an aperture or around objects. Diffraction occurs when the size of the aperture or obstacle is of the same order of magnitude as the wavelength of the incident wave. For very small aperture sizes, the vast majority of the wave is blocked. in case of large apertures the wave passes by or through the obstacle without any significant diffraction.

7 0

3 years ago

A miniature quadcopter is located at x = -2.25 m and y, - 5.70 matt - 0 and moves with an average velocity having components Vv,

kupik [55]

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 x-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 y-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 x- and y-components of the copter's position vector after t = 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

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

so if any of these are incorrect, you should make the appropriate adjustments to the work above.

8 0

3 years ago