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3 years ago

MULTIPLE CHOICE PLEASE HELP QUICK!!!!!!

Physics

2 answers:

Natasha_Volkova [10]3 years ago

6 0

Answer:

934 m/s and it is faster in hydrogen than in air

Explanation:

Speed of sound in gas is given by the formula

$v = \sqrt{\frac{\gamma RT}{M}}$

here we know for hydrogen

M = 2 g/mol

R = 8.31 J/mol k

$\gamma = 1.4$

T = 0 degree C = 273 K

now we will have

$v = \sqrt{\frac{(1.4)(8.31)(273)}{0.002}}$

$v_{hydrogen} = 1260.2 m/s$

Now for air we know that

M = 29 g/mol

R = 8.31 J/mol k

$\gamma = 1.4$

T = 30 degree C = 303 K

now we will have

$v = \sqrt{\frac{(1.4)(8.31)(303)}{0.029}}$

$v_{air} = 348.6 m/s$

so the difference in the speed is given as

$\Delta v = v_{hydrogen} - v_{air}$

$\Delta v = 1260.2 - 348.6 = 911.4 m/s$

so the closest answer is 934 m/s and it is faster in hydrogen than in air

vekshin13 years ago

5 0

i would choose a for 945 and a for faster

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A charge of -8.00 nC is spread uniformly over the surface of one face of a nonconducting disk of radius 1.05 cm.

gavmur [86]

Answer:

(a) $E = -1.02 \times 10^5~N/C$

(b) $E = -9.7 \times 10^4~N/C$

Explanation:

(a) The electric field for a point charge is given by the following formula:

$\vec{E} = \frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}\^r$

Since this formula is valid for point charges, we have to choose an infinitesimal area, da, from the disk. Then we will calculate the E-field (dE) created by this small area using the above formula, then we will integrate over the entire disk to find the E-field created by the disk.

$dE = \frac{1}{4\pi\epsilon_0}\frac{dQ}{(\sqrt{z^2 + r^2})^2}$

Here, z = 0.025 m. And r is the distance of the infinitesimal area from the axis. dQ is the charge of the small area, and should be written in terms of the given variables.

In cylindrical coordinates, da = r dr dθ. So,

$\frac{Q}{\pi R^2} = \frac{dQ}{da}\\\frac{Q}{\pi R^2} = \frac{dQ}{rdrd\theta}\\dQ = \frac{Qrdrd\theta}{\pi R^2}$

Hence, dE is now:

$dE = \frac{1}{4\pi\epsilon_0}\frac{Q}{\pi R^2}\frac{rdrd\theta}{z^2 + r^2}$

The surface integral over the disk can now be taken, but there is one more thing to be considered. This dE is a vector quantity, and it needs to be separated its components.

It has two components, one in the vertical direction and another in the horizontal direction. By symmetry, the horizontal components cancel out each other in the end (since it is a disk, each horizontal vector has an equal but opposite counterpart), so only the vertical component should be considered.

Let us denote the angle between dE and the horizontal axis as α. This angle can be found by the geometry of the triangle formed by dE, vertical axis of the disk, and horizontal plane. So,

$\sin(\alpha) = \frac{z}{\sqrt{z^2 + r^2}}$

Therefore, vertical component of dE now becomes

$dE_z = \frac{1}{4\pi\epsilon_0}\frac{Q}{\pi R^2}\frac{rdrd\theta}{z^2 + r^2}\frac{z}{\sqrt{z^2+r^2}} = \frac{1}{4\pi\epsilon_0}\frac{Qz}{\pi R^2}\frac{rdrd\theta}{(z^2+r^2)^{3/2}}\\E_z = \frac{1}{4\pi\epsilon_0}\frac{Qz}{\pi R^2}\int\limits^{2\pi}_0 \int\limits^R_0 {\frac{rdrd\theta}{(z^2+r^2)^{3/2}}} = \frac{1}{4\pi\epsilon_0}\frac{Qz}{\pi R^2} 2\pi(\frac{1}{z} - \frac{1}{\sqrt{z^2+R^2}})$

Substituting the parameters, z = 0.025 m, Q = - 8 x 10^(-9) C, and R = 0.0105 m, yields the final result:

$E_z = \frac{1}{2\epsilon_0}\frac{Qz}{\pi R^2}(\frac{1}{z} - \frac{1}{\sqrt{z^2+R^2}}) = -1.02 \times 10^5~N/C$

(b) We will have a similar approach, but a simpler integral.

$dE = \frac{1}{4\pi\epsilon_0}\frac{dQ}{z^2 + R^2}\\\frac{Q}{2\pi R} = \frac{dQ}{Rd\theta}\\dQ = \frac{Qd\theta}{2\pi}\\dE = \frac{1}{4\pi\epsilon_0}\frac{Qd\theta}{2\pi(z^2 + R^2)}\\dE_z = \frac{1}{4\pi\epsilon_0}\frac{Qd\theta}{2\pi(z^2 + R^2)}\frac{z}{\sqrt{z^2+R^2}} = \frac{1}{4\pi\epsilon_0}\frac{Qzd\theta}{2\pi(z^2 + R^2)^{3/2}}\\E_z = \frac{1}{4\pi\epsilon_0}\frac{Qz}{2\pi(z^2 + R^2)^{3/2}}\int\limits^{2\pi}_0 {} \, d\theta = \frac{1}{4\pi\epsilon_0}\frac{Qz}{2\pi(z^2 + R^2)^{3/2}}2\pi$

$E_z = \frac{1}{4\pi\epsilon_0}\frac{Qz}{(z^2 + R^2)^{3/2}} = -9.07\times 10^4~N/C$

Note that, in this case the source object is a one dimensional hoop rather than a two dimensional disk.

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3 years ago

Tick (3) the correct statement about electrostatic charges.

Radda [10]

Answer:

similar type of electric charges attract one another

I think this is a coorect staement

6 0

3 years ago

Read 2 more answers

Two strings on a musical instrument are tuned to play at 196 hz (g) and 523 hz (c). (a) what are the first two overtones for eac

Tems11 [23]

(a) first two overtones for each string:
The first string has a fundamental frequency of 196 Hz. The n-th overtone corresponds to the (n+1)-th harmonic, which can be found by using
$f_n = n f_1$
where f1 is the fundamental frequency.

So, the first overtone (2nd harmonic) of the string is
$f_2 = 2 f_1 = 2 \cdot 196 Hz = 392 Hz$
while the second overtone (3rd harmonic) is
$f_3 = 3 f_1 = 3 \cdot 196 Hz = 588 Hz$

Similarly, for the second string with fundamental frequency $f_1 = 523 Hz$ , the first overtone is
$f_2 = 2 f_1 = 2 \cdot 523 Hz = 1046 Hz$
and the second overtone is
$f_3 = 3 f_1 = 3 \cdot 523 Hz = 1569 Hz$

(b) The fundamental frequency of a string is given by
$f= \frac{1}{2L} \sqrt{ \frac{T}{\mu} }$
where L is the string length, T the tension, and $\mu = m/L$ is the mass per unit of length. This part of the problem says that the tension T and the length L of the string are the same, while the masses are different (let's calle them $m_{196}$ , the mass of the string of frequency 196 Hz, and $m_{523}$ , the mass of the string of frequency 523 Hz.
The ratio between the fundamental frequencies of the two strings is therefore:
$\frac{523 Hz}{196 Hz} = \frac{ \frac{1}{2L} \sqrt{ \frac{T}{m_{523}/L} } }{\frac{1}{2L} \sqrt{ \frac{T}{m_{196}/L} }}$
and since L and T simplify in the equation, we can find the ratio between the two masses:
$\frac{m_{196}}{m_{523}}=( \frac{523 Hz}{196 Hz} )^2 = 7.1$

(c) Now the tension T and the mass per unit of length $\mu$ is the same for the strings, while the lengths are different (let's call them $L_{196}$ and $L_{523}$ ). Let's write again the ratio between the two fundamental frequencies
$\frac{523 Hz}{196 Hz}= \frac{ \frac{1}{2L_{523}} \sqrt{ \frac{T}{\mu} } }{\frac{1}{2L_{196}} \sqrt{ \frac{T}{\mu} }}$
And since T and $\mu$ simplify, we get the ratio between the two lengths:
$\frac{L_{196}}{L_{523}}= \frac{523 Hz}{196 Hz}=2.67$

(d) Now the masses m and the lenghts L are the same, while the tensions are different (let's call them $T_{196}$ and $T_{523}$ . Let's write again the ratio of the frequencies:
$\frac{523 Hz}{196 Hz}= \frac{ \frac{1}{2L} \sqrt{ \frac{T_{523}}{m/L} } }{\frac{1}{2L} \sqrt{ \frac{T_{196}}{m/L} }}$
Now m and L simplify, and we get the ratio between the two tensions:
$\frac{T_{196}}{T_{523}}=( \frac{196 Hz}{523 Hz} )^2=0.14$

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