Density: g/mL, kg/cubic meter
Volume: L, teaspoon
Mass: g, MeV/sq. C
Answer:
![= - 5.65\times 10^{-2} J](https://tex.z-dn.net/?f=%3D%20-%205.65%5Ctimes%2010%5E%7B-2%7D%20J)
Explanation:
Given data:
L =2.00 *10^4 m
d = 18*10^4 m
M = 18 *10^6 kg
m_1 = 8*10^6 kg
Gravitational energy is given as
![U =- G \frac{m_1 m_2}{r}](https://tex.z-dn.net/?f=U%20%3D-%20G%20%5Cfrac%7Bm_1%20m_2%7D%7Br%7D)
mass per unit length is given as
![\sigma = \frac{M}{L} = \frac{18 \times 10^6}{2\times 10^4 m} = 900 kg/m](https://tex.z-dn.net/?f=%5Csigma%20%3D%20%5Cfrac%7BM%7D%7BL%7D%20%3D%20%5Cfrac%7B18%20%5Ctimes%2010%5E6%7D%7B2%5Ctimes%2010%5E4%20m%7D%20%3D%20900%20kg%2Fm)
calculating potential energy
![dU ==-G\int_{16\times 10^4}^{18\times 10^4} \frac{m_1 *dm}{r}](https://tex.z-dn.net/?f=dU%20%3D%3D-G%5Cint_%7B16%5Ctimes%2010%5E4%7D%5E%7B18%5Ctimes%2010%5E4%7D%20%5Cfrac%7Bm_1%20%2Adm%7D%7Br%7D)
![=-G\int_{16\times 10^4}^{18\times 10^4} \frac{m_1 *\sigma dr}{r}](https://tex.z-dn.net/?f=%3D-G%5Cint_%7B16%5Ctimes%2010%5E4%7D%5E%7B18%5Ctimes%2010%5E4%7D%20%5Cfrac%7Bm_1%20%2A%5Csigma%20dr%7D%7Br%7D)
![=-G*m_1*\sigma\int_{16\times 10^4}^{18\times 10^4} \frac{dr}{r}](https://tex.z-dn.net/?f=%3D-G%2Am_1%2A%5Csigma%5Cint_%7B16%5Ctimes%2010%5E4%7D%5E%7B18%5Ctimes%2010%5E4%7D%20%5Cfrac%7Bdr%7D%7Br%7D)
![=-G*m_1*\sigma \left | ln r \right |_{16\times 10^4}^{18\times 10^4}](https://tex.z-dn.net/?f=%3D-G%2Am_1%2A%5Csigma%20%5Cleft%20%7C%20ln%20r%20%5Cright%20%7C_%7B16%5Ctimes%2010%5E4%7D%5E%7B18%5Ctimes%2010%5E4%7D)
![=-G*m_1*\sigma ln(1.125)](https://tex.z-dn.net/?f=%3D-G%2Am_1%2A%5Csigma%20ln%281.125%29)
![=-6.673 \times 10^{-11}*8*10^6*900*ln(1.125)](https://tex.z-dn.net/?f=%3D-6.673%20%5Ctimes%2010%5E%7B-11%7D%2A8%2A10%5E6%2A900%2Aln%281.125%29)
![= - 5.65\times 10^{-2} J](https://tex.z-dn.net/?f=%3D%20-%205.65%5Ctimes%2010%5E%7B-2%7D%20J)
Answer:
2.6 m
Explanation:
The work done by the bird is given by
![W=Fd](https://tex.z-dn.net/?f=W%3DFd)
where
F is the force exerted
d is the distance covered
In this problem, we know:
is the work
is the force
Solving the equation for d, we find the distance covered by the bird:
![d=\frac{W}{F}=\frac{8.8\cdot 10^{-4}}{3.4\cdot 10^{-4}}=2.6 m](https://tex.z-dn.net/?f=d%3D%5Cfrac%7BW%7D%7BF%7D%3D%5Cfrac%7B8.8%5Ccdot%2010%5E%7B-4%7D%7D%7B3.4%5Ccdot%2010%5E%7B-4%7D%7D%3D2.6%20m)
To develop this problem it is necessary to apply the oscillation frequency-related concepts specifically in string or pipe close at both ends or open at both ends.
By definition the oscillation frequency is defined as
![f = n\frac{v}{2L}](https://tex.z-dn.net/?f=f%20%3D%20n%5Cfrac%7Bv%7D%7B2L%7D)
Where
v = speed of sound
L = Length of the pipe
n = any integer which represent the number of repetition of the spectrum (n)1,2,3...)(Number of harmonic)
Re-arrange to find L,
![f = n\frac{v}{2L}\\L = \frac{nv}{2f}](https://tex.z-dn.net/?f=f%20%3D%20n%5Cfrac%7Bv%7D%7B2L%7D%5C%5CL%20%3D%20%5Cfrac%7Bnv%7D%7B2f%7D)
The radius between the two frequencies would be 4 to 5,
![\frac{528Hz}{660Hz}= \frac{4}{5}](https://tex.z-dn.net/?f=%5Cfrac%7B528Hz%7D%7B660Hz%7D%3D%20%5Cfrac%7B4%7D%7B5%7D)
![4:5](https://tex.z-dn.net/?f=4%3A5)
Therefore the frequencies are in the ratio of natural numbers. That is
![4f = 528\\f = \frac{528}{4}\\f = 132Hz](https://tex.z-dn.net/?f=4f%20%3D%20528%5C%5Cf%20%3D%20%5Cfrac%7B528%7D%7B4%7D%5C%5Cf%20%3D%20132Hz)
Here f represents the fundamental frequency.
Now using the expression to calculate the Length we have
![L = \frac{nv}{2f}\\L = \frac{(1)343m/s}{2(132)}\\L = 1.29m](https://tex.z-dn.net/?f=L%20%3D%20%5Cfrac%7Bnv%7D%7B2f%7D%5C%5CL%20%3D%20%5Cfrac%7B%281%29343m%2Fs%7D%7B2%28132%29%7D%5C%5CL%20%3D%201.29m)
Therefore the length of the pipe is 1.3m
For the second harmonic n=2, then
![L = \frac{nv}{2f}\\L = \frac{(2)343m/s}{2(132)}\\L = 2.59m](https://tex.z-dn.net/?f=L%20%3D%20%5Cfrac%7Bnv%7D%7B2f%7D%5C%5CL%20%3D%20%5Cfrac%7B%282%29343m%2Fs%7D%7B2%28132%29%7D%5C%5CL%20%3D%202.59m)
Therefore the length of the pipe in the second harmonic is 2.6m