Answer:
KE₂ = 6000 J
Explanation:
Given that
Potential energy at top U₁= 7000 J
Potential energy at bottom U₂= 1000 J
The kinetic energy at top ,KE₁= 0 J
Lets take kinetic energy at bottom level = KE₂
Now from energy conservation
U₁+ KE₁= U₂+ KE₂
Now by putting the values
U₁+ KE₁= U₂+ KE₂
7000+ 0 = 1000+ KE₂
KE₂ = 7000 - 1000 J
KE₂ = 6000 J
Therefore the kinetic energy at bottom is 6000 J.
Answer: ![71.16\ Hz](https://tex.z-dn.net/?f=71.16%5C%20Hz)
Explanation:
Given
Capacitance ![C=100\ \mu F](https://tex.z-dn.net/?f=C%3D100%5C%20%5Cmu%20F)
Resistance ![R=500\ \Omega](https://tex.z-dn.net/?f=R%3D500%5C%20%5COmega)
Inductance ![L=50\ mH](https://tex.z-dn.net/?f=L%3D50%5C%20mH)
In LCR circuit, current is maximum at resonance frequency i.e.
![X_L=X_C\ \text{and}\ \omega_o=\dfrac{1}{\sqrt{LC}}](https://tex.z-dn.net/?f=X_L%3DX_C%5C%20%5Ctext%7Band%7D%5C%20%5Comega_o%3D%5Cdfrac%7B1%7D%7B%5Csqrt%7BLC%7D%7D)
Insert the values
![\Rightarrow \omega_o=\dfrac{1}{\sqrt{50\times 10^{-3}\times 100\times 10^{-6}}}\\\\\Rightarrow \omega_o=\dfrac{1}{\sqrt{5}\times 10^{-3}}\\\\\Rightarrow \omega_o=0.447\times 10^{3}](https://tex.z-dn.net/?f=%5CRightarrow%20%5Comega_o%3D%5Cdfrac%7B1%7D%7B%5Csqrt%7B50%5Ctimes%2010%5E%7B-3%7D%5Ctimes%20100%5Ctimes%2010%5E%7B-6%7D%7D%7D%5C%5C%5C%5C%5CRightarrow%20%5Comega_o%3D%5Cdfrac%7B1%7D%7B%5Csqrt%7B5%7D%5Ctimes%2010%5E%7B-3%7D%7D%5C%5C%5C%5C%5CRightarrow%20%5Comega_o%3D0.447%5Ctimes%2010%5E%7B3%7D)
Also, frequency is given by
![\Rightarrow 2\pi f=\omega_o\\\\\Rightarrow f=\frac{\omega_o}{2\pi}](https://tex.z-dn.net/?f=%5CRightarrow%202%5Cpi%20f%3D%5Comega_o%5C%5C%5C%5C%5CRightarrow%20f%3D%5Cfrac%7B%5Comega_o%7D%7B2%5Cpi%7D)
![\Rightarrow f=\dfrac{1}{2\pi}\times 0.447\times 10^3\\\\\Rightarrow f=71.16\ Hz](https://tex.z-dn.net/?f=%5CRightarrow%20f%3D%5Cdfrac%7B1%7D%7B2%5Cpi%7D%5Ctimes%200.447%5Ctimes%2010%5E3%5C%5C%5C%5C%5CRightarrow%20f%3D71.16%5C%20Hz)
Half of the moon is illuminated.
This is unclear. What are the objects? Is it a balloon? A rubber ball?