Since sound is a mechanical wave it needs a ...... to travel through

Physics

1 answer:

Leni [432]3 years ago

6 0

Answer:

medium

Examples:

A medium is a thing that the sound wave would travel through, like sound waves traveling through air. A mechanical wave cannot travel without a medium, like if the wave were trying to travel through a vacuum.

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A potter spins his wheel at 0.98 rev/s. The wheel has a mass of 4.2 kg and a radius of 0.35 m. He drops a chunk of clay of 2.9 k

Bad White [126]

Answer:

$v_{f,w} = 1.791\,\frac{m}{s}$ , $v_{f,c} = 0.972\,\frac{m}{s}$

Explanation:

The situation can be modelled by applying the Principle of Angular Momentum Conservation:

$I_{w} \cdot \omega_{o} = (I_{w} + I_{c})\cdot \omega_{f}$

The final angular speed is:

$\omega_{f} = \frac{I_{w}}{I_{w}+I_{c}}\cdot \omega_{o}$

$\omega_{f} = \left(\frac{\frac{1}{2}\cdot (4.2\,kg)\cdot (0.35\,m)^{2} }{\frac{1}{2}\cdot (4.2\,kg)\cdot (0.35\,m)^{2} + \frac{1}{2}\cdot (2.9\,kg)\cdot (0.19\,m)^{2}}\right)\cdot (0.98\,\frac{rev}{s} )\cdot \left(\frac{2\pi\,rad}{1\,rev} \right)$

$\omega_{f} \approx 5.116\,\frac{rad}{s}$

The tangential velocities of the wheel and the clay are, respectively:

$v_{f, w} = (0.35\,m)\cdot (5.116\,\frac{rad}{s} )$

$v_{f,w} = 1.791\,\frac{m}{s}$

$v_{f, c} = (0.19\,m) \cdot (5.116\,\frac{rad}{s} )$

$v_{f,c} = 0.972\,\frac{m}{s}$

5 0

3 years ago

A listener triples his distance from a source that emits sound uniformly in all directions. by how many decibels does the sound

Novay_Z [31]

Sound intensity goes inversely as the square of the distance from the source. This is easy to understand if you think of the total sound energy spread out over the surface of a sphere with the source at the center. The surface area of the sphere goes as r^2, so the power density goes inversely as such.

The decibel change is simply

Change = 10 log10( (1/3)^2 ) = -9.54 dB

8 0

3 years ago

uppose that the terminal speed of a particular sky diver is 150 km/h in the spread-eagle position and 320 km/h in the nosedive p

AysviL [449]

Answer:

4.55

Explanation:

The terminal speed of a diver is given by:

$v_t=\sqrt{\frac{2mg}{C\rho A} } \\\\Where\ m=mass\ of \ driver,d=acceleration\ due\ to\ gravity,C=drag\ \\coefficient,A=cross\ sectional\ Area.\\\\Therefore:\\\\A=\frac{2mg}{C \rho v_t^2} \\\\For\ area\ with\ terminal\ speed\ in\ spread\ angle\ position(v_s):\\\\A_s=\frac{2mg}{C \rho v_s^2} \\\\For\ area\ with\ terminal\ speed\ in\ nose\ dive\ position(v_n):\\\\A_n=\frac{2mg}{C \rho v_n^2}\\\\Therefore\ since\ g,m,C,\rho\ are\ constant:\\\\$

$\frac{A_s}{A_n}= \frac{\frac{2mg}{C \rho v_s^2}}{\frac{2mg}{C \rho v_n^2}}\\\\\frac{A_s}{A_n}= \frac{v_n}{v_s} \\\\v_n=320\ km/h,v_s=150\ km/h\\\\\frac{A_s}{A_n}=\frac{320^2}{150^2} =4.55$