False. An Electromagnetic induction is the production of an electromotive force across an electrical conductor in a changing magnetic field.
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)
The work done to pull the sled up to the hill is given by
![W=Fd](https://tex.z-dn.net/?f=W%3DFd)
where
F is the intensity of the force
d is the distance where the force is applied.
In our problem, the work done is
![W=1900 J](https://tex.z-dn.net/?f=W%3D1900%20J)
and the distance through which the force is applied is
![d=400 m](https://tex.z-dn.net/?f=d%3D400%20m)
, so we can calculate the average force by re-arranging the previous equation and by using these data:
Answer:
W = 1418.9 J = 1.418 KJ
Explanation:
In order to find the work done by the pull force applied by Karla, we need to can use the formula of work done. This formula tells us that work done on a body is the product of the distance covered by the object with the component of force applied in the direction of that displacement:
W = F.d
W = Fd Cosθ
where,
W = Work Done = ?
F = Force = 151 N
d = distance covered = 10 m
θ = Angle with horizontal = 20°
Therefore,
W = (151 N)(10 m) Cos 20°
<u>W = 1418.9 J = 1.418 KJ</u>
temperature decreases at higher altitudes because as air rises the pressure decreases.