Mercury does not wet glass because of the cohesive forces within the drops are stronger than the adhesive forces between the drops and glass.
Answer:
a) 4.49Hz
b) 0.536kg
c) 2.57s
Explanation:
This problem can be solved by using the equation for he position and velocity of an object in a mass-string system:
![x=Acos(\omega t)\\\\v=-\omega Asin(\omega t)\\\\a=-\omega^2Acos(\omega t)](https://tex.z-dn.net/?f=x%3DAcos%28%5Comega%20t%29%5C%5C%5C%5Cv%3D-%5Comega%20Asin%28%5Comega%20t%29%5C%5C%5C%5Ca%3D-%5Comega%5E2Acos%28%5Comega%20t%29)
for some time t you have:
x=0.134m
v=-12.1m/s
a=-107m/s^2
If you divide the first equation and the third equation, you can calculate w:
![\frac{x}{a}=\frac{Acos(\omega t)}{-\omega^2 Acos(\omega t)}\\\\\omega=\sqrt{-\frac{a}{x}}=\sqrt{-\frac{-107m/s^2}{0.134m}}=28.25\frac{rad}{s}](https://tex.z-dn.net/?f=%5Cfrac%7Bx%7D%7Ba%7D%3D%5Cfrac%7BAcos%28%5Comega%20t%29%7D%7B-%5Comega%5E2%20Acos%28%5Comega%20t%29%7D%5C%5C%5C%5C%5Comega%3D%5Csqrt%7B-%5Cfrac%7Ba%7D%7Bx%7D%7D%3D%5Csqrt%7B-%5Cfrac%7B-107m%2Fs%5E2%7D%7B0.134m%7D%7D%3D28.25%5Cfrac%7Brad%7D%7Bs%7D)
with this value you can compute the frequency:
a)
![f=\frac{\omega}{2\pi}=\frac{28.25rad/s}{2\pi}=4.49Hz](https://tex.z-dn.net/?f=f%3D%5Cfrac%7B%5Comega%7D%7B2%5Cpi%7D%3D%5Cfrac%7B28.25rad%2Fs%7D%7B2%5Cpi%7D%3D4.49Hz)
b)
the mass of the block is given by the formula:
![f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}\\\\m=\frac{k}{4\pi^2f^2}=\frac{427N/m}{(4\pi^2)(4.49Hz)^2}=0.536kg](https://tex.z-dn.net/?f=f%3D%5Cfrac%7B1%7D%7B2%5Cpi%7D%5Csqrt%7B%5Cfrac%7Bk%7D%7Bm%7D%7D%5C%5C%5C%5Cm%3D%5Cfrac%7Bk%7D%7B4%5Cpi%5E2f%5E2%7D%3D%5Cfrac%7B427N%2Fm%7D%7B%284%5Cpi%5E2%29%284.49Hz%29%5E2%7D%3D0.536kg)
c) to find the amplitude of the motion you need to know the time t. This can computed by dividing the equation for v with the equation for x and taking the arctan:
![\frac{v}{x}=-\omega tan(\omega t)\\\\t=\frac{1}{\omega}arctan(-\frac{v}{x\omega })=\frac{1}{28.25rad/s}arctan(-\frac{-12.1m/s}{(0.134m)(28.25rad/s)})=2.57s](https://tex.z-dn.net/?f=%5Cfrac%7Bv%7D%7Bx%7D%3D-%5Comega%20tan%28%5Comega%20t%29%5C%5C%5C%5Ct%3D%5Cfrac%7B1%7D%7B%5Comega%7Darctan%28-%5Cfrac%7Bv%7D%7Bx%5Comega%20%7D%29%3D%5Cfrac%7B1%7D%7B28.25rad%2Fs%7Darctan%28-%5Cfrac%7B-12.1m%2Fs%7D%7B%280.134m%29%2828.25rad%2Fs%29%7D%29%3D2.57s)
Finally, the amplitude is:
![x=Acos(\omega t)\\\\A=\frac{0.134m}{cos(28.25rad/s*2.57s )}=0.45m](https://tex.z-dn.net/?f=x%3DAcos%28%5Comega%20t%29%5C%5C%5C%5CA%3D%5Cfrac%7B0.134m%7D%7Bcos%2828.25rad%2Fs%2A2.57s%20%29%7D%3D0.45m)
Answer:
If he's 500 kg and 1 kg = 10 N then it would be 5,000 N
Explanation:
The surface area is unimportant and they'll use it to trick you.
The answer is D. <span>d. Event 4 may cause a hurricane and Event 2 may cause a tornado.
Event 4, the difference in pressure cause by this will resulted in the air to start moving uncontrollably, which ending up in Hurricane
Event 2, this will create energy needed for thunderstorm to be formed which could lead to Tornado</span>
sorry accidentally posted here.