Answer:
the spring constant k = ![5.409*10^4 \ N/m](https://tex.z-dn.net/?f=5.409%2A10%5E4%20%5C%20N%2Fm)
the value for the damping constant ![\\ \\b = 1.518 *10^3 \ kg/s](https://tex.z-dn.net/?f=%5C%5C%20%5C%5Cb%20%3D%201.518%20%2A10%5E3%20%5C%20kg%2Fs)
Explanation:
From Hooke's Law
![F = kx\\\\k =\frac{F}{x}\\\\where \ F = mg\\\\k = \frac{mg}{x}\\\\given \ that:\\\\mass \ of \ each \ wheel = 425 \ kg\\\\x = 7.7cm = 0.077 m\\\\g = 9.8 \ m/s^2\\\\Then;\\\\k = \frac{425 \ kg * 9.8 \ m/s^2}{0.077 \ m}\\\\k = 5.409*10^4 \ N/m](https://tex.z-dn.net/?f=F%20%3D%20kx%5C%5C%5C%5Ck%20%3D%5Cfrac%7BF%7D%7Bx%7D%5C%5C%5C%5Cwhere%20%5C%20F%20%3D%20mg%5C%5C%5C%5Ck%20%3D%20%5Cfrac%7Bmg%7D%7Bx%7D%5C%5C%5C%5Cgiven%20%5C%20that%3A%5C%5C%5C%5Cmass%20%5C%20of%20%5C%20each%20%5C%20wheel%20%3D%20425%20%5C%20kg%5C%5C%5C%5Cx%20%3D%207.7cm%20%3D%200.077%20m%5C%5C%5C%5Cg%20%3D%209.8%20%5C%20m%2Fs%5E2%5C%5C%5C%5CThen%3B%5C%5C%5C%5Ck%20%3D%20%5Cfrac%7B425%20%5C%20kg%20%2A%209.8%20%5C%20m%2Fs%5E2%7D%7B0.077%20%5C%20m%7D%5C%5C%5C%5Ck%20%3D%205.409%2A10%5E4%20%5C%20N%2Fm)
Thus; the spring constant k = ![5.409*10^4 \ N/m](https://tex.z-dn.net/?f=5.409%2A10%5E4%20%5C%20N%2Fm)
The amplitude is decreasing 37% during one period of the motion
![e^{\frac{-bT}{2m}}= \frac{37}{100}\\\\e^{\frac{-bT}{2m}}= 0.37\\\\\frac{-bT}{2m} = In(0.37)\\\\\frac{-bT}{2m} = -0.9943\\\\b = \frac{2m(0.9943)}{T}\\\\b = \frac{2m(0.9943)}{\frac{2 \pi}{\omega}}\\\\b = \frac{m(0.9943) \ ( \omega) )}{ \pi}](https://tex.z-dn.net/?f=e%5E%7B%5Cfrac%7B-bT%7D%7B2m%7D%7D%3D%20%5Cfrac%7B37%7D%7B100%7D%5C%5C%5C%5Ce%5E%7B%5Cfrac%7B-bT%7D%7B2m%7D%7D%3D%200.37%5C%5C%5C%5C%5Cfrac%7B-bT%7D%7B2m%7D%20%3D%20In%280.37%29%5C%5C%5C%5C%5Cfrac%7B-bT%7D%7B2m%7D%20%3D%20-0.9943%5C%5C%5C%5Cb%20%3D%20%5Cfrac%7B2m%280.9943%29%7D%7BT%7D%5C%5C%5C%5Cb%20%3D%20%5Cfrac%7B2m%280.9943%29%7D%7B%5Cfrac%7B2%20%5Cpi%7D%7B%5Comega%7D%7D%5C%5C%5C%5Cb%20%3D%20%5Cfrac%7Bm%280.9943%29%20%5C%20%28%20%5Comega%29%20%29%7D%7B%20%5Cpi%7D)
![b = \frac{m(0.9943)(\sqrt{\frac{k}{m})}}{\pi}\\\\b = \frac{425*(0.9943)(\sqrt{\frac{5.409*10^4}{425}) } }{3.14}\\\\b = 1518.24 \ kg/s\\\\b = 1.518 *10^3 \ kg/s](https://tex.z-dn.net/?f=b%20%3D%20%5Cfrac%7Bm%280.9943%29%28%5Csqrt%7B%5Cfrac%7Bk%7D%7Bm%7D%29%7D%7D%7B%5Cpi%7D%5C%5C%5C%5Cb%20%3D%20%5Cfrac%7B425%2A%280.9943%29%28%5Csqrt%7B%5Cfrac%7B5.409%2A10%5E4%7D%7B425%7D%29%20%7D%20%20%20%20%7D%7B3.14%7D%5C%5C%5C%5Cb%20%3D%201518.24%20%5C%20kg%2Fs%5C%5C%5C%5Cb%20%3D%201.518%20%2A10%5E3%20%5C%20kg%2Fs)
Therefore; the value for the damping constant ![\\ \\b = 1.518 *10^3 \ kg/s](https://tex.z-dn.net/?f=%5C%5C%20%5C%5Cb%20%3D%201.518%20%2A10%5E3%20%5C%20kg%2Fs)
Answer:
375 ms
Explanation:
the frequency of metronome , f = 160 beats per minute
f = 160 /60 beats per sec
f = 2.67 beats /s
the period of a single beat , T = 1/f
T = 1/2.67 s
T = 0.375 s = 375 ms
the period of a single beat is 375 ms
To solve this problem, it will be necessary to apply the concepts related to the fundamental resonance frequency in a closed organ pipe.
This is mathematically given as
![f_n (2n+1)(\frac{v}{4L})](https://tex.z-dn.net/?f=f_n%20%282n%2B1%29%28%5Cfrac%7Bv%7D%7B4L%7D%29)
For fundamental frequency n is 0, then,
![f_0 = \frac{v}{4L}](https://tex.z-dn.net/?f=f_0%20%3D%20%5Cfrac%7Bv%7D%7B4L%7D)
When,
v = Velocity of sound
L = Length,
Rearranging to find the velocity,
![v = f_0 (4L)](https://tex.z-dn.net/?f=v%20%3D%20f_0%20%284L%29)
![v = (80Hz)(4)(1.3m)](https://tex.z-dn.net/?f=v%20%3D%20%2880Hz%29%284%29%281.3m%29)
![v = 416m/s](https://tex.z-dn.net/?f=v%20%3D%20416m%2Fs)
Therefore the speed of sound in this gas is 416m/s
Ionization energy, according to <span>chem.libretexts.org,</span><span> is the quantity of </span>energy<span> that an isolated, gaseous atom in the ground electronic state must absorb to discharge an electron, resulting in a cation. This </span>energy<span> is usually expressed in kJ/mol, or the amount of </span>energy<span> it takes for all the atoms in a mole to lose one electron each.</span>
Answer:
![U(3)=-43J](https://tex.z-dn.net/?f=U%283%29%3D-43J)
Explanation:
Potential energy is minus the integral of Fdx. Doing the integration yields:
![U=\int\limits {6.0x+12}\, dx](https://tex.z-dn.net/?f=U%3D%5Cint%5Climits%20%7B6.0x%2B12%7D%5C%2C%20dx)
![U=-3x^2-12x+C](https://tex.z-dn.net/?f=U%3D-3x%5E2-12x%2BC)
![U(0)=20J](https://tex.z-dn.net/?f=U%280%29%3D20J)
so
![U(0)=-3(0)^2+12(0)+C=20](https://tex.z-dn.net/?f=U%280%29%3D-3%280%29%5E2%2B12%280%29%2BC%3D20)
![C=20](https://tex.z-dn.net/?f=C%3D20)
Now for x=3.0m
![U(3)=-3*(3)^2-12(3)+20](https://tex.z-dn.net/?f=U%283%29%3D-3%2A%283%29%5E2-12%283%29%2B20)
![U(3)=-43J](https://tex.z-dn.net/?f=U%283%29%3D-43J)