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kati45 [8]
2 years ago
11

When a 12 V battery is connected to a 6 uF capacitor, how much energy is stored

Physics
1 answer:
adoni [48]2 years ago
4 0

The energy stored in a capacitor is

E = (1/2) · (capacitance) · (voltage)²

E = (1/2) · (6 x 10⁻⁶ F) · (12 V)²

E = (3 x 10⁻⁶ F) · (144 V²)

<em>E = 4.32 x 10⁻⁴ Joule</em>

(That's 0.000432 of a Joule)

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Explanation:

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A submarine periscope uses two totally reflecting 45-45-90 prisms with total internal reflection on the sides adjacent to the 45
Likurg_2 [28]

Answer

Given,

Periscope uses 45-45-90 prisms with total internal reflection adjacent to 45°.

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refractive index of glass, n_g = 1.52

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3 years ago
The school bag of four students A,B,C,D measures 9kg, 2800gm, 2kg and 8000gm respectively. Whose bag is the lightest
inna [77]

Answer:

Student C

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2 years ago
What is the orbital period of a spacecraft in a low orbit near the surface of mars? The radius of mars is 3.4×106m.
valkas [14]
<h2>Answer: 56.718 min</h2>

Explanation:

According to the Third Kepler’s Law of Planetary motion<em> </em><em>“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”. </em>

In other words, this law states a relation between the orbital period T of a body (moon, planet, satellite) orbiting a greater body in space with the size a of its orbit.

This Law is originally expressed as follows:

T^{2}=\frac{4\pi^{2}}{GM}a^{3}   (1)

Where;

G is the Gravitational Constant and its value is 6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}

M=6.39(10)^{23}kg is the mass of Mars

a=3.4(10)^{6}m  is the semimajor axis of the orbit the spacecraft describes around Mars (assuming it is a <u>circular orbit </u>and a <u>low orbit near the surface </u>as well, the semimajor axis is equal to the radius of the orbit)

If we want to find the period, we have to express equation (1) as written below and substitute all the values:

T=\sqrt{\frac{4\pi^{2}}{GM}a^{3}}    (2)

T=\sqrt{\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(6.39(10)^{23}kg)}(3.4(10)^{6}m)^{3}}    (3)

T=\sqrt{11581157.44 s^{2}}    (4)

Finally:

T=3403.1099s=56.718min    This is the orbital period of a spacecraft in a low orbit near the surface of mars

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