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RUDIKE [14]
4 years ago
5

A tetrahedron has an equilateral triangle base with 25.0-cm-long edges and three equilateral triangle sides. The base is paralle

l to the ground, and a vertical uniform electric field of strength 260N/C passes upward through the tetrahedron.
What is the electric flux through the base?







What is the electric flux through each of the three sides?
Physics
1 answer:
Jet001 [13]4 years ago
6 0

Answer:

-7.03645640575 Nm²/C

2.34548546858 Nm²/C

Explanation:

S = Surface area

Electric flux through the base is given by

\phi_b=-\int Eds\\\Rightarrow \phi_b=-ES\\\Rightarrow \phi_b=-260\dfrac{\sqrt{3}}{4}0.25^2\\\Rightarrow \phi_b=-7.03645640575\ Nm^2/C

The electric flux through the base is -7.03645640575 Nm²/C

Inside the tetrahedron there is no charge so total flux will be zero

\phi_b+3\phi_s=0\\\Rightarrow \phi_s=-\dfrac{\phi_b}{3}\\\Rightarrow \phi_s=-\dfrac{-7.03645640575}{3}\\\Rightarrow \phi_s=2.34548546858\ Nm^2/C

The electric flux through the sides is 2.34548546858 Nm²/C

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<h3>Introduction</h3>

Hi ! Here I will discuss about the angular frequency or what is also often called the angular velocity because it has the same unit dimensions. <u>Angular frequency occurs, when an object vibrates (either moving harmoniously / oscillating or moving in a circle)</u>. Angular frequency can be roughly interpreted as the magnitude of the change in angle (in units of rad) per unit time. So, based on this understanding, the angular frequency can be calculated using the equation :

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With the following condition :

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  • t = interval of the time (s)

<h3>Problem Solving</h3>

We know that :

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  • t = interval of the time = 54.9 s.

What was asked :

  • \sf{\omega} = angular frequency = ... rad/s

Step by step :

\sf{\omega = \frac{\theta}{t}}

\sf{\omega = \frac{2,000 \pi}{54.9}}

\boxed{\sf{\omega \approx 36.43 \pi \: rad/s}}

<h3>Conclusion :</h3>

So, the angular frequency of the blades approximately 36.43π rad/s.

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