Answer: c
Explanation:
Analytical methods
Answer:
![=\frac{1/3}{5/6} = 0.4](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%2F3%7D%7B5%2F6%7D%20%3D%200.4)
Explanation:
Moment of inertia of given shell![= \frac{2}{3} MR^2](https://tex.z-dn.net/?f=%20%3D%20%5Cfrac%7B2%7D%7B3%7D%20MR%5E2)
where
M represent sphere mass
R -sphere radius
we know linear speed is given as ![v = r\omega](https://tex.z-dn.net/?f=v%20%3D%20r%5Comega)
translational ![K.E = \frac{1}{2} mv^2 = \frac{1}{2} m(r\omega)^2](https://tex.z-dn.net/?f=K.E%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20mv%5E2%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20m%28r%5Comega%29%5E2)
rotational ![K.E = \frac{1}{2} I \omega^2 = \frac{1}{2} \frac{2}{3} MR^2 \omega^2](https://tex.z-dn.net/?f=K.E%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20I%20%5Comega%5E2%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%5Cfrac%7B2%7D%7B3%7D%20MR%5E2%20%5Comega%5E2)
total kinetic energy will be
![K.E = \frac{1}{2} m(r\omega)^2 + \frac{1}{2} \frac{2}{3} MR^2 \omega^2](https://tex.z-dn.net/?f=K.E%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20m%28r%5Comega%29%5E2%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20%5Cfrac%7B2%7D%7B3%7D%20MR%5E2%20%5Comega%5E2)
![K.E =\frac{5}{6} MR^2 \omega^2](https://tex.z-dn.net/?f=K.E%20%3D%5Cfrac%7B5%7D%7B6%7D%20MR%5E2%20%5Comega%5E2)
fraction of rotaional to total K.E
![=\frac{1/3}{5/6} = 0.4](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%2F3%7D%7B5%2F6%7D%20%3D%200.4)
The energy from the sun that reaches the corn is about two billionths.
Answer:
x = -1.20 m
y = -1.12 m
Explanation:
as we know that four masses and their position is given as
5.0 kg (0, 0)
2.9 kg (0, 3.2)
4 kg (2.5, 0)
8.3 kg (x, y)
As we know that the formula of center of gravity is given as
![x_{cm} = \frac{m_1 x_1 + m_2x_2 + m_3x_3 + m_4x_4}{m_1 + m_2 + m_3 + m_4}](https://tex.z-dn.net/?f=x_%7Bcm%7D%20%3D%20%5Cfrac%7Bm_1%20x_1%20%2B%20m_2x_2%20%2B%20m_3x_3%20%2B%20m_4x_4%7D%7Bm_1%20%2B%20m_2%20%2B%20m_3%20%2B%20m_4%7D)
![0 = \frac{5(0) + 2.9(0) + 4(2.5) + 8.3 x}{5 + 2.9 + 4 + 8.3}](https://tex.z-dn.net/?f=%200%20%3D%20%5Cfrac%7B5%280%29%20%2B%202.9%280%29%20%2B%204%282.5%29%20%2B%208.3%20x%7D%7B5%20%2B%202.9%20%2B%204%20%2B%208.3%7D)
![10 + 8.3 x = 0](https://tex.z-dn.net/?f=10%20%2B%208.3%20x%20%3D%200)
![x = -1.20 m](https://tex.z-dn.net/?f=x%20%3D%20-1.20%20m)
Similarly for y direction we have
![y_{cm} = \frac{m_1 y_1 + m_2y_2 + m_3y_3 + m_4y_4}{m_1 + m_2 + m_3 + m_4}](https://tex.z-dn.net/?f=y_%7Bcm%7D%20%3D%20%5Cfrac%7Bm_1%20y_1%20%2B%20m_2y_2%20%2B%20m_3y_3%20%2B%20m_4y_4%7D%7Bm_1%20%2B%20m_2%20%2B%20m_3%20%2B%20m_4%7D)
![0 = \frac{5(0) + 2.9(3.2) + 4(0) + 8.3 y}{5 + 2.9 + 4 + 8.3}](https://tex.z-dn.net/?f=%200%20%3D%20%5Cfrac%7B5%280%29%20%2B%202.9%283.2%29%20%2B%204%280%29%20%2B%208.3%20y%7D%7B5%20%2B%202.9%20%2B%204%20%2B%208.3%7D)
![9.28 + 8.3 x = 0](https://tex.z-dn.net/?f=9.28%20%2B%208.3%20x%20%3D%200)
![x = -1.12 m](https://tex.z-dn.net/?f=x%20%3D%20-1.12%20m)
What's now called "Conventional current" is thought of as the flow of positive charge, from the battery's positive terminal to its negative one.
But it turns out that positive charges don't flow. The physical flow of charge is the flow of electrons. They come out of the battery's negative terminal, and carry negative charge around the circuit to the battery's positive one.