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AfilCa [17]
1 year ago
13

A student has made the statement that the electric flux through one half of a Gaussian surface is always equal and opposite to t

he flux through the other half of the Gaussian surface. This is:_______.
a. never true.

b. never false.

c. true whenever enclosed charge is symmetrically located at a center point, or on a center line or centrally placed plane

d. true whenever no charge is enclosed within the Gaussian surface.

e. true only when no charge is enclosed within the Gaussian surface.
Physics
1 answer:
nata0808 [166]1 year ago
6 0

Answer:

E.true only when no charge is enclosed within the Gaussian surface.

Explanation:

Because Gauss’s law states that the net flux of an electric field in a closed surface is directly proportional to the enclosed electric charge.

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Answer:

B. 4 m/s

Explanation:

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Distinguish between the terms strength, power, and endurance as they are used in weightlifting ​
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The internal shear force V at a certain section of a steel beam is 80 kN, and the moment of inertia is 64,900,000 . Determine th
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Here is the complete question

The internal shear force V at a certain section of a steel beam is 80 kN, and the moment of inertia is 64,900,000 . Determine the horizontal shear stress at point H, which is located L  = 20 mm below the centriod

The missing image which is the remaining part of this question is attached in the image below.

Answer:

The horizontal shear stress at point H is  \mathbf{\tau_H \approx  42.604 \ N/mm^2}

Explanation:

Given that :

The internal shear force V  =  80 kN = 80 × 10³ N

The moment of inertia = 64,900,000

The length = 20 mm below the centriod

The horizontal shear stress  \tau can be calculated by using the equation:

\tau = \dfrac{VQ}{Ib}

where;

Q = moment of area above or below the point H

b = thickness of the beam = 10  mm

From the centroid ;

Q = Q_1 + Q_{2}

Q = A_1y_1 + A_{2}y_{2}  

Q = ( ( 70 × 10) × (55) + ( 210 × 15) (90 + 15/2) ) mm³

Q = ( ( 700) × (55) + ( 3150 ) ( 97.5)  ) mm³

Q = ( 38500 +  307125 ) mm³

Q = 345625 mm³

\tau_H = \dfrac{VQ}{Ib}

\tau_H = \dfrac{80*10^3  * 345625}{64900000*10 }

\tau_H = \dfrac{2.765*10^{10}}{649000000 }

\tau_H = 42.60400616 \ N/mm^2

\mathbf{\tau_H \approx  42.604 \ N/mm^2}

The horizontal shear stress at point H is  \mathbf{\tau_H \approx  42.604 \ N/mm^2}

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