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)
Since the ball was not moving before it let Aiden's hand, the formula used to calculate the acceleration is
![a = \frac{v}{t}](https://tex.z-dn.net/?f=a%20%3D%20%20%5Cfrac%7Bv%7D%7Bt%7D%20)
, where a is acceleration, v is velocity and t is the time. We put them in the formula and get
![a = \frac{49}{0.1} \\ a = \frac{490}{1} \\ a = 490](https://tex.z-dn.net/?f=a%20%3D%20%20%5Cfrac%7B49%7D%7B0.1%7D%20%20%5C%5C%20a%20%3D%20%20%5Cfrac%7B490%7D%7B1%7D%20%20%5C%5C%20a%20%3D%20490)
The acceleration is 490 m/s^2
Answer:
![1 \times 10 { - }^{9}](https://tex.z-dn.net/?f=1%20%5Ctimes%2010%20%7B%20-%20%7D%5E%7B9%7D%20)
cubic metre or 1e-9
Explanation:
•By division. Number of cubic millimetre divided(/) by 1000000000, equal(=): Number of cubic metre.
•By multiplication. 83 mm3(s) * 1.0E-9 = 8.3E-8 m3(s)
Answer:
Umm that's a personal question. All u have to do is say when have u pushed your personal limits....... Ummm one for me is when i had to try out for a select soccer and that is past my comfort zone.
Explanation:
The energy bar eaten by Sheila has chemical energy locked up inside it. This chemical energy is converted to mechanical energy in form of potential and kinetic energy and this in turn is converted to heat energy as the run progresses. Thus, the energy changes are: chemical energy to mechanical energy [kinetic and potential] and finally to heat energy.