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klio [65]
1 year ago
10

What is the electric flux passing through a gaussian surface that surrounds a 0.075 c point charge?

Physics
1 answer:
Valentin [98]1 year ago
7 0

An 0.075 c point charge's surrounding gaussian surface's electric flux  is 8.5 ×10⁹Nm²/C.

<h3>Which is a gaussian surface?</h3>

In three dimensions, the Gaussian surface is referred to as a closed surface where the flux of a vector field may be determined. The gravitational field, the electric field, or the magnetic field are all examples of these vector fields.

<h3>Why is a Gaussian surface drawn?</h3>

We build a fictitious Gaussian surface around the supplied surface when the surface of which an electric field or flux must be established is asymmetrical or the surface area is challenging to obtain, such as the surface area for an infinitely long wire or plane. The Gaussian surface is said to utilize symmetry the best.

When a charge q is surrounded by a gaussian surface, the electric flux that passes through it is

\phi=q / \epsilon 0= 0.075 / ( 8.85 ×10⁻¹²)

= 8.5 ×10⁹Nm²/C.

To know more about Gaussian Surface visit:

brainly.com/question/13003278

#SPJ4

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Compare the time period of two simple pendulums of length 4m and 16m at a place.
Vlad1618 [11]

Answer:

the period of the 16 m pendulum is twice the period of the 4 m pendulum

Explanation:

Recall that the period (T) of a pendulum of length (L)  is defined as:

T=2\,\pi\,\sqrt{ \frac{L}{g} }

where "g" is the local acceleration of gravity.

SInce both pendulums are at the same place, "g" is the same for both, and when we compare the two periods, we get:

T_1=2\,\pi\,\sqrt{\frac{4}{g} } \\T_2=2\,\pi\,\sqrt{\frac{16}{g} } \\ \\\frac{T_2}{T_1} =\sqrt{\frac{16}{4} } =2

therefore the period of the 16 m pendulum is twice the period of the 4 m pendulum.

5 0
3 years ago
A two slit pattern is viewed on a screen 1.00m from the slits if the two third-order minima are 22.0 cm apart what is the width
Bingel [31]

Answer:

4.4 cm

Explanation:

Given:

Distance of the screen from the slit, D = 1 m

Distance between two third order interference minimas, x = 22 cm

Let's say, minima occurs at:

x_n = (n + \frac{1}{2}) \frac{wL}{d}

We have:

2x_2 = 2(2 + \frac{1}{2}) * \frac{w*22}{d}

Calculating further for the width of the central bright fringe, we have:

\frac{w}{d} = \frac{22}{5}

= 4.4 cm

Note: w in representswavelength

8 0
3 years ago
The image shows street lights powered by solar panels. Which sequence shows the energy transformations taking place in these lig
yawa3891 [41]

the answer is C. for plato

8 0
3 years ago
Read 2 more answers
Determine the Mutual Inductance per unit length between two long solenoids, one inside the other, whose radii are r1 and r2 (r2
Triss [41]

Answer:

M' = μ₀n₁n₂πr₂²

Explanation:

Since r₂ < r₁ the mutual inductance M = N₂Ф₂₁/i₁ where N₂ = number of turns of solenoid 2 = n₂l where n₂ = number of turns per unit length of solenoid 2 and l = length of solenoid, Ф₂₁ = flux in solenoid 2 due to magnetic field in solenoid 1 = B₁A₂ where B₁ = magnetic field due to solenoid 1 = μ₀n₁i₁ where μ₀ = permeability of free space, n₁ = number of turns per unit length of solenoid 1 and i₁ = current in solenoid 1. A₂ = area of solenoid 2 = πr₂² where r₂ = radius of solenoid 2.

So, M = N₂Ф₂₁/i₁

substituting the values of the variables into the equation, we have

M = N₂Ф₂₁/i₁

M = N₂B₁A₂/i₁

M = n₂lμ₀n₁i₁πr₂²/i₁

M = lμ₀n₁n₂πr₂²

So, the mutual inductance per unit length is M' = M/l = μ₀n₁n₂πr₂²

M' = μ₀n₁n₂πr₂²

3 0
3 years ago
A balloon filled with helium gas at 20°C occupies 4.91 L at 1.00 atm. The balloon is immersed in liquid nitrogen at -196°C, whil
mrs_skeptik [129]

Answer:

0.25 L

Explanation:

P_1 = Initial pressure = 1 atm

T_1 = Initial Temperature = 20 °C

V_1 = Initial volume = 4.91 L

P_2 = Final pressure = 5.2 atm

T_2 = Final Temperature = -196 °C

V_2= Final volume

From ideal gas law we have

\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\\\Rightarrow V_2=\dfrac{P_1V_1T_2}{T_1P_2}\\\Rightarrow P_2=\dfrac{1\times 4.91(273.15-196)}{(20+273.15)\times 5.2}\\\Rightarrow V_2=0.24849\ L\approx 0.25\ L

The pressure experienced by the balloon is 0.25 L

7 0
3 years ago
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