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3 years ago

What is the electric flux ΦΦPhi through each of the six faces of the cube?

Physics

1 answer:

denpristay [2]3 years ago

7 0

Flux through each of the six faces of the cube: $\frac{q}{6\epsilon_0}$

Explanation:

In this problem, a charge $q$ is placed at the center of the cube.

We can apply Gauss Law, which states that the flux through the surface of the cube is equal to the charge contained within the cube divided by the vacuum permittivity; mathematically:

$\Phi_T=\frac{q}{\epsilon_0}$

where

$q$ is the charge

$\epsilon_0$ is the vacuum permittivity

Here we want to find the flux through each of the six faces of the cube.

By simmetry, we can say that the 6 faces are identical: therefore, the flux through each of them must be the same. This means that the flux through each faces is 1/6 of the total flux through the total surface, therefore:

$\Phi = \frac{1}{6}\Phi = \frac{q}{6\epsilon_0}$

Learn more about electric fields:

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Two men are standing on a frictionless ice surface holding opposite ends of a rope. One man (mass = 80 kg) pulls on the rope wit

ankoles [38]

Answer:

The acceleration of man 1 and 2 is $3.125\ m/s^2$ and $4.167\ m/s^2$ .

Explanation:

Mass of man 1, m₁ = 80 kg

Mass of man 2, m₂ = 60 kg

One man pulls on the rope with a force of 250 N.

Let a₁ is acceleration of man 1,

F = m₁a₁

$a_1=\dfrac{F}{m_1}\\\\a_1=\dfrac{250}{80}\\\\a_1=3.125\ m/s^2$

Let a₂ is acceleration of man 1,

F = m₂a₂

$a_2=\dfrac{F}{m_2}\\\\a_2=\dfrac{250}{60}\\\\a_2=4.167\ m/s^2$

So, the acceleration of man 1 and 2 is $3.125\ m/s^2$ and $4.167\ m/s^2$ .

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3 years ago

Whiat is quantum numbers describes the size and energy of an orbital?

Levart [38]

Answer:

The answer is the principal Quantum number (n)

Explanation:

The principal quantum number is one of the four quantum numbers associated with an atom.

It is denoted by a number n=1,2,3,4 etc

It tells both size (directly) and energy (indirectly) of an orbital.

When n=1 means it is the closest to the nucleus and is the smallest orbital and with increase in principal quantum number, it depicts that size of the orbital is increasing.

It tells the energy of the orbital as well as smaller number means less distance from nucleus and having less energy. Since electrons requires to absorb energy to jump into higher orbitals making n=2,3,4 etc. Thus electrons in the orbitals with higher n number indicates higher energy orbitals.

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3 years ago

A 4.67-g bullet is moving horizontally with a velocity of +357 m/s, where the sign + indicates that it is moving to the right (s

Leni [432]

Answer:

(a)0.531m/s

(b)0.00169

Explanation:

We are given that

Mass of bullet, m=4.67 g= $4.67\times 10^{-3} kg$

1 kg =1000 g

Speed of bullet, v=357m/s

Mass of block 1, $m_1=1177g=1.177kg$

Mass of block 2, $m_2=1626 g=1.626 kg$

Velocity of block 1, $v_1=0.681m/s$

(a)

Let velocity of the second block after the bullet imbeds itself=v2

Using conservation of momentum

Initial momentum=Final momentum

$mv=m_1v_1+(m+m_2)v_2$

$4.67\times 10^{-3}\times 357+1.177(0)+1.626(0)=1.177\times 0.681+(4.67\times 10^{-3}+1.626)v_2$

$1.66719=0.801537+1.63067v_2$

$1.66719-0.801537=1.63067v_2$

$0.865653=1.63067v_2$

$v_2=\frac{0.865653}{1.63067}$

$v_2=0.531m/s$

Hence, the velocity of the second block after the bullet imbeds itself=0.531m/s

(b)Initial kinetic energy before collision

$K_i=\frac{1}{2}mv^2$

$k_i=\frac{1}{2}(4.67\times 10^{-3}\times (357)^2)$

$k_i=297.59 J$

Final kinetic energy after collision

$K_f=\frac{1}{2}m_1v^2_1+\frac{1}{2}(m+m_2)v^2_2$

$K_f=\frac{1}{2}(1.177)(0.681)^2+\frac{1}{2}(4.67\times 10^{-3}+1.626)(0.531)^2$

$K_f=0.5028 J$

Now, he ratio of the total kinetic energy after the collision to that before the collision

= $\frac{k_f}{k_i}=\frac{0.5028}{297.59}$

=0.00169

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Answer:

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