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Delvig [45]
3 years ago
9

Why does the earth bulge at the equator?

Physics
1 answer:
sattari [20]3 years ago
4 0

centrifugal force is a fictitious force. What is happening is that since the earth itself is not a rigid body it will deform when under motion. Although gravity attempts to make the earth spherical, as it is rotating the earth deforms, in such away that it flattens to become an oblique spheroid. This happens as the material at the equator must have a net resultant centripetal force (not centrifugal) which causes its position of equilibrium from the center of the earth to be further away than at the poles as they do not have this force as they are not rotating around the center of mass.

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A parallel-plate vacuum capacitor has 8.60 J of energy stored in it. The separation between the plates is 3.80 mm . If the separ
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Answer:

Part a)

E = \frac{8.60}{2.62} = 3.28 J

Part b)

E = 2.62(8.60) = 22.5 J

Explanation:

As we know that the energy of capacitor when it is not connected to potential source is given as

U = \frac{Q^2}{2C}

As we know that initial energy is given as

8.60 = \frac{Q^2}{2C}

now we know that capacitance of parallel plate capacitor is given as

C = \frac{\epsilon_0A}{d}

now the new capacitance when distance is changed from 3.80 mm to 1.45 mm

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C' = \frac{C(3.80)}{1.45}

C' = 2.62 C

Now the new energy of the capacitor is given as

E = \frac{Q^2}{2(2.62C)}

E = \frac{8.60}{2.62} = 3.28 J

Part b)

Now if the voltage difference between the plates of capacitor is given constant

now the energy energy of capacitor is

U = \frac{1}{2}CV^2

8.60 = \frac{1}{2}CV^2

now when capacitance is changed to new value then new energy is given as

E = \frac{1}{2}C'V^2

E = \frac{1}{2}(2.62C)V^2

E = 2.62(8.60) = 22.5 J

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