Answer: hertz (Hz) i think :)
Answer:
Explanation:
Given that;
horizontal circle at a rate of 2.33 revolutions per second
the magnetic field of the Earth is 0.500 gauss
the baton is 60.1 cm in length.
the magnetic field is oriented at 14.42°
we wil get the area due to rotation of radius of baton is
![\Delta A = \frac{1}{2} \Delta \theta R^2](https://tex.z-dn.net/?f=%5CDelta%20A%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%5CDelta%20%5Ctheta%20R%5E2)
The formula for the induced emf is
![E = \frac{\Delta \phi}{\Delta t}](https://tex.z-dn.net/?f=E%20%3D%20%5Cfrac%7B%5CDelta%20%20%5Cphi%7D%7B%5CDelta%20%20t%7D)
![\phi = \texttt {magnetic flux}](https://tex.z-dn.net/?f=%5Cphi%20%20%3D%20%5Ctexttt%20%7Bmagnetic%20flux%7D)
![E=\frac{\Delta (BA) }{\Delta t}](https://tex.z-dn.net/?f=E%3D%5Cfrac%7B%5CDelta%20%28BA%29%20%7D%7B%5CDelta%20%20t%7D)
![=B\frac{\Delta A}{\Delta t}](https://tex.z-dn.net/?f=%3DB%5Cfrac%7B%5CDelta%20%20A%7D%7B%5CDelta%20%20t%7D)
B is the magnetic field strength
substitute
![\texttt {substitute}\ \frac{1}{2} \Delta \theta R^2 \ \ for \Delta A](https://tex.z-dn.net/?f=%5Ctexttt%20%7Bsubstitute%7D%5C%20%20%5Cfrac%7B1%7D%7B2%7D%20%5CDelta%20%5Ctheta%20R%5E2%20%5C%20%5C%20for%20%5CDelta%20%20A)
![E=B\frac{(\Delta \theta R^3/2)}{\Delta t} \\\\=\frac{1}{2} BR^2\omega](https://tex.z-dn.net/?f=E%3DB%5Cfrac%7B%28%5CDelta%20%20%5Ctheta%20R%5E3%2F2%29%7D%7B%5CDelta%20%20t%7D%20%5C%5C%5C%5C%3D%5Cfrac%7B1%7D%7B2%7D%20BR%5E2%5Comega)
The magnetic field of the earth is oriented at 14.42
![\omega =2.33\\\\L=60.1c,\\\\\theta=14.42\\\\B=0.5](https://tex.z-dn.net/?f=%5Comega%20%3D2.33%5C%5C%5C%5CL%3D60.1c%2C%5C%5C%5C%5C%5Ctheta%3D14.42%5C%5C%5C%5CB%3D0.5)
we plug in the values in the equation above
so, the induce EMF will be
![E=\frac{1}{2} \times (B\sin \theta)R^2\omega\\\\E=\frac{1}{2} \times (B\sin \theta)(\frac{L}{2} )\omega](https://tex.z-dn.net/?f=E%3D%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%20%28B%5Csin%20%5Ctheta%29R%5E2%5Comega%5C%5C%5C%5CE%3D%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%20%28B%5Csin%20%5Ctheta%29%28%5Cfrac%7BL%7D%7B2%7D%20%29%5Comega)
![=\frac{1}{2} \times0.5gauss\times\frac{0.0001T}{1gauss} \times\sin 14.42\times(\frac{60.1\times10^-^2m}{2} )^2(2.33rev/s)(\frac{2\pi rad}{1rev} )\\\\=2.5\times10^-^5\times0.2490\times0.0903\times14.63982\\\\=2.5\times10^-^5\times0.32917\\\\=8.229\times10^-^6V](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes0.5gauss%5Ctimes%5Cfrac%7B0.0001T%7D%7B1gauss%7D%20%5Ctimes%5Csin%2014.42%5Ctimes%28%5Cfrac%7B60.1%5Ctimes10%5E-%5E2m%7D%7B2%7D%20%29%5E2%282.33rev%2Fs%29%28%5Cfrac%7B2%5Cpi%20rad%7D%7B1rev%7D%20%29%5C%5C%5C%5C%3D2.5%5Ctimes10%5E-%5E5%5Ctimes0.2490%5Ctimes0.0903%5Ctimes14.63982%5C%5C%5C%5C%3D2.5%5Ctimes10%5E-%5E5%5Ctimes0.32917%5C%5C%5C%5C%3D8.229%5Ctimes10%5E-%5E6V)
The correct answer is, A) Straight line motion
I took the quiz
To create the shapes, stars are arranged on a piece of cardboard in the desired configuration. If the stars are placed in a smiley face pattern on the cardboard, for example, they will explode into a smiley face in the sky. In fact, you may see several smiley faces in the sky at one time.
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