# Tag Info

4

John Rennie has provided an exact mathematical treatment of the equations behind the calculation of the speed of sound. I don't want to detract from that treatment, and of course the Wikipedia articles we both draw from provide a broader treatment; but an intuitive understanding of the 'why' has been equally helpful for me, in the past. The following is my ...

0

Your question should more accurately have been, "Why does sound travel faster in solid iron than in liquid mercury even though mercury has higher density?" Were the question phrased that way, the answer would be more obvious. At temperatures at which both metals are liquid or both metals are solid, sound travels faster in the denser metal.

3

The square of the sound velocity is proportional to the ratio of an elastic modulus to the mass density of the material.The reason why the sound velocity is usually larger in solids than in liquids and usually larger in liquids than in gases is because of the elastics constants of the material. What determines the elastic constants of a material is the ...

28

The speed of sound in a liquid is given by: $$v = \sqrt{\frac{K}{\rho}}$$ where $K$ is the bulk modulus and $\rho$ is the density. The bulk modulus of mercury is $2.85 \times 10^{10}$ Pa and the density is $13534$ kg/m$^3$, so the equation gives $v = 1451$ m/sec. The speed of sound in solids is given by: $$v = \sqrt{\frac{K + \tfrac{4}{3}G}{\rho}}$$ ...

6

"The speed of sound is variable and depends on the properties of the substance through which the wave is traveling. In solids, the speed of transverse (or shear) waves depend on the shear deformation under shear stress (called the shear modulus), and the density of the medium. Longitudinal (or compression) waves in solids depend on the same two factors with ...

0

Your answer for (b)(ii) is a little too simplistic. The idea is right, but the formula for I_encl(r_1 < r < r_2) is the same as your answer for (a), except with r_2 replaced by r: I_encl(r_1 < r < r_2) = π·α·(r⁴-r_1⁴)/2 . Therefore B(r_1 < r < r_2) = µ_0·α/4·(r⁴-r_1⁴)/r . I_encl(r > r_2) is just the total current flowing through the wire, ...

-1

We start from point (5) above where things started to go wrong: $\frac {dV_c}{ dt}= - \frac {dV_b}{ dt} \tag{5}$ $\frac {dV_c} {dt} = m_c / (\frac {d\rho_c}{dt} ) = - \frac {m_c } {0.1} \tag{6}$ \begin{align} \int_{V_c,0}^{V_c,1} {dV_c} = - \frac {m_c } {0.1} \int_0^t dt \tag{7} \end{align} \begin{align} V_{c,1}-V_{c,0} = - \frac {m_c } ... 1 In your 5th "Let" statement "\frac {d\rho_c}{ dt} = 0.1kg/m^3/sec$the rate pressure changes in the large container", "pressure" should be "density" and there is a sign error. Your equations 1-4 are correct, but it is important to determine what is known and what is unknown and analyze how many independent equations and unknowns you have. Also, mass of ... 1 Normally when compressing a gas the temperature increases. If you assume adiabatic compression, the law is$PV^\gamma=k$, where$\gamma=\frac {C_P}{C_V}$is the ratio of specific heats and is usually about$1.4$for air. Then, as shown here$\frac {T_2}{T_1}=\left(\frac {P_2}{P_1}\right)^{\gamma-\frac 1\gamma}\$ This assumes you don't leak heat to the ...

-1

D=m/v, (with rounded numbers) Sphere for volume= (4/3)(pi)(radius^3) r (cm)=(1.0x10^-13)cm mass (g)= (1.7x10^-24) =(1.7x10^-24)/((4/3)pi(1.0x10^-13)^3) =(1.7x10^-24)/(4.1887x10^39) = 4.0585x10^14

-1

http://www.amazon.com/s/ref=nb_sb_noss_1?url=search-alias%3Dindustrial&field-keywords=mercury&rh=n%3A16310091%2Ck%3Amercury is an excellent source (the one I was talking about above)

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