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attashe74 [19]
2 years ago
6

If this combination of capacitors is attached to a 9 v battery how much energy will be stored on the plates 25 15

Physics
1 answer:
Vlad1618 [11]2 years ago
4 0

The energy stored on the plates will be 378.5 J. The combination of capacitors is in series.

<h3>What is a capacitor?</h3>

A capacitor is a device that can store electrical energy. It is a two-conductor configuration separated by an insulating medium that carries charges of equal size and opposite sign.

An electric insulator or vacuum, such as glass, paper, air, or a semiconductor termed a dielectric, can be used as the non-conductive zone.

The complete question is;

"If this combination of capacitors is attached to a 9.00 V battery; how much energy will be 25 μF 15 μF stored on the plates?"

If the two capacitors are in series the equivalent capacitance will be;

\rm \frac {1}{C} = \frac{1}{C_1}+\frac{1}{C_2} \\\\ \rm \frac {1}{C} = \frac{1}{25}+\frac{1}{15} \\\\ C = 9.375

Energy will be stored on the plates;

\rm E = \frac{1}{2}CV^2 \\\\ E= \frac{1}{2} \times 9.375 \mu F \times ( 9 \ V)^2\\\\E = 378.5 \ J

Hence the energy stored on the plates will be 378.5 J.

To learn more about the capacitor, refer to the link: brainly.com/question/14048432.

#SPJ1

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One electron collides elastically with a second electron initially at rest. After the collision, the radii of their trajectories
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Answer:

114.92749 keV

Explanation:

r = Radius of trajectory

m = Mass of electron = 9.11\times 10^{-31}\ kg

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The centripetal force and the magnetic forces are conserved

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K=K_1+K_2\\\Rightarrow K=\frac{1}{2}mv_1^2+\frac{1}{2}mv_2^2\\\Rightarrow K=\frac{1}{2}m(v_1^2+v_2^2)\\\Rightarrow K=\frac{1}{2}\times 9.11\times 10^{-31}(77277716.79473^2+185466520.30735^2)\\\Rightarrow K=1.83884\times 10^{-14}\ J

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1\ J=\frac{1}{1.6\times 10^{-19}}\ eV

1.83884\times 10^{-14}\ J=1.83884\times 10^{-14}\times \frac{1}{1.6\times 10^{-19}}\ eV\\ =114927.49\ ev=114.92749\ keV

The energy of incident electron is 114.92749 keV

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