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ladessa [460]
3 years ago
5

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

4 0
3 years ago
A substance is soluble in water. It is added into the water and dissolves. More solute continues being added until it reaches a
Sophie [7]

Answer:

option D

Explanation:

also to add.... there exists a solubility equilibrium between solid solute and the solution.

4 0
4 years ago
The drawings show three charges that have the same magnitude but may have different signs. In all cases the distance d between t
Radda [10]

Answer:

a)F = 6816.5680 N

b) F' = 0 N

c)F''=28195.5N

Explanation:

Magnitude of charges M=1.6\muC

Distance d=2.6

Generally the equation for  Net Force is mathematically given by

For First Drawing

F =\frac{ k q^2}{d^2}+\frac{k q^2}{d^2}  

F =\frac{2* 9*10^9* (1.6*10^-6)^2}{ (2.6*10^-3^2}

F = 6816.5680 N

For second Drawing

F' =\frac{ k q^2}{d^2 }-\frac{ k q^2}{ d^2}  

F' = 0 N

For Third Drawing

F'' =\frac{ k q^2}{d^2} * \sqrt (2)

F'' = 9*10^9* (6.4*10^-6)^2 * \frac{\sqrt(2)}{(4.3*10^-3)^2}

F''=28195.5N

4 0
3 years ago
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