Current will be
![I=\dfrac{V_{rms}}{Z}=\dfrac{V_{rms}}{\sqrt{ R^{2}+(X_{C}-X_{L})^{2}}}\\where~X_{c}=\dfrac{1}{j.\omega .C}~and~X_{L}=j.\omega.L~where~\omega=2.\pi f~and~f=60Hz](https://tex.z-dn.net/?f=%20I%3D%5Cdfrac%7BV_%7Brms%7D%7D%7BZ%7D%3D%5Cdfrac%7BV_%7Brms%7D%7D%7B%5Csqrt%7B%20R%5E%7B2%7D%2B%28X_%7BC%7D-X_%7BL%7D%29%5E%7B2%7D%7D%7D%5C%5Cwhere~X_%7Bc%7D%3D%5Cdfrac%7B1%7D%7Bj.%5Comega%20.C%7D~and~X_%7BL%7D%3Dj.%5Comega.L~where~%5Comega%3D2.%5Cpi%20f~and~f%3D60Hz)
now just pluf in the values and Voila..
Answer:
Explanation:
a ) starting from rest , so u = o and initial kinetic energy = 0 .
Let mass of the skier = m
Kinetic energy gained = potential energy lost
= mgh = mg l sinθ
= m x 9.8 x 70 x sin 30
= 343 m
Total kinetic energy at the base = 343 m + 0 = 343 m .
b )
In this case initial kinetic energy = 1/2 m v²
= .5 x m x 2.5²
= 3.125 m
Total kinetic energy at the base
= 3.125 m + 343 m
= 346.125 m
c ) It is not surprising as energy gained due to gravitational force by the earth is enormous . So component of energy gained due to gravitational force far exceeds the initial kinetic energy . Still in a competitive event , the fractional initial kinetic energy may be the deciding factor .
Each year is a different anount of hurricanes but on the Pacific Ocean there is about 14 hurricanes (that's form).
Hurricanes rotate clockwise. The Saffir-Simpson scale, it's measured on a scale 1-5 property damage and flooding along the coast. Hurricanes get their names from meteorologist a long time ago from their communication about them being more effectively and to tell whether to stay more farther away from the hurricane. The calmest winds in the hurricane is the, Eye. Which the Eye is also called the, cyclone. The strongest winds are in the right-front side of the hurricanes center. Hope this helped! Sorry that I couldn't get everything.
Answer:
Speed of 0.08 kg mass when it will reach to the bottom position is 1.94 m/s
Explanation:
When rod is released from rest then due to unbalanced torque about the hinge the system will rotate
Now moment of inertia of the system is given as
![I = \frac{ML^2}{12} + \frac{m_1L^2}{4} + \frac{m_2L^2}{4}](https://tex.z-dn.net/?f=I%20%3D%20%5Cfrac%7BML%5E2%7D%7B12%7D%20%2B%20%5Cfrac%7Bm_1L%5E2%7D%7B4%7D%20%2B%20%5Cfrac%7Bm_2L%5E2%7D%7B4%7D)
now we have
![M = 0.120 kg](https://tex.z-dn.net/?f=M%20%3D%200.120%20kg)
![m_1 = 0.02 kg](https://tex.z-dn.net/?f=m_1%20%3D%200.02%20kg)
![m_3 = 0.08 kg](https://tex.z-dn.net/?f=m_3%20%3D%200.08%20kg)
now we have
![I = \frac{0.120(0.90)^2}{12} + \frac{0.02(0.90)^2}{4} + \frac{0.08(0.90)^2}{4}](https://tex.z-dn.net/?f=I%20%3D%20%5Cfrac%7B0.120%280.90%29%5E2%7D%7B12%7D%20%2B%20%5Cfrac%7B0.02%280.90%29%5E2%7D%7B4%7D%20%2B%20%5Cfrac%7B0.08%280.90%29%5E2%7D%7B4%7D)
so we have
![I = 8.1 \times 10^[-3} + 4.05 \times 10^[-3} + 0.0162](https://tex.z-dn.net/?f=I%20%3D%208.1%20%5Ctimes%2010%5E%5B-3%7D%20%2B%204.05%20%5Ctimes%2010%5E%5B-3%7D%20%2B%200.0162)
![I = 0.02835](https://tex.z-dn.net/?f=I%20%3D%200.02835)
now by energy conservation we can say work done by gravity must be equal to change in kinetic energy
so we have
![\frac{1}{2}I\omega^2 = m_1g \frac{L}{2} - m_2 g\frac{L}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7DI%5Comega%5E2%20%3D%20m_1g%20%5Cfrac%7BL%7D%7B2%7D%20-%20m_2%20g%5Cfrac%7BL%7D%7B2%7D)
![\frac{1}{2}(0.02835)\omega^2 = (0.08 - 0.02)(9.81)(0.45)](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%280.02835%29%5Comega%5E2%20%3D%20%280.08%20-%200.02%29%289.81%29%280.45%29)
![\omega = 4.32 rad/s](https://tex.z-dn.net/?f=%5Comega%20%3D%204.32%20rad%2Fs)
Now speed of 0.08 kg mass when it reaches to bottom point is given as
![v = \omega \frac{L}{2}](https://tex.z-dn.net/?f=v%20%3D%20%5Comega%20%5Cfrac%7BL%7D%7B2%7D)
![v = 4.32 (0.45)](https://tex.z-dn.net/?f=v%20%3D%204.32%20%280.45%29)
![v = 1.94 m/s](https://tex.z-dn.net/?f=v%20%3D%201.94%20m%2Fs)