Answer:
Bar graph
Pie graph
Line Graph
Explanation:
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Kinetic energy is the energy of motion.
The formula for kinetic energy is given as
KE = (0.5) m v²
where m = mass of object , v = speed of object.
an object having some speed and mass will have kinetic energy while an object at rest will not have any kinetic energy since the speed of object at rest is zero.
Since at rest speed is zero. an object has kinetic energy only when it is in motion.
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)