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allsm [11]
3 years ago
13

Gauss's Law You are in the presence of a uniform electric field of strength E that is pointed in the x direction. You have 2 Gau

ssian surfaces; one is a square of side length L, and the other is a circle of diameter L. Which has the larger electric flux value?
Physics
1 answer:
BartSMP [9]3 years ago
7 0

Answer:

the flow is the same on both surfaces

Explanation:

Gauss's law says that the electric field flow through a Gaussian surface is equal to the charge inside the surface divided by the electric permittivity constant, the equation is

          Φ =∫ E . dA = q / ε₀

The point is the scalar product and the bold ones indicate vector,

Let us examine the situations that give us, the elective field is uniform so that the field lines that cross the surface of the square are constant and perpendicular to two sides, therefore in flow on one side it is equal to the flow on the other side, but of negative sign; so the net flow is zero, in the two sides there are 90º between the field lines and the normal to the surface (dA) so the flow is zero

Now let's analyze the circular surface in this case the angles between the field that is uniform, lines are parallel, and the normal circle changes at each point, but by the symmetry of the circle the angle that there is an area at the top is the same of the surface at the bottom connected by a diameter, the net flow in each pair of points is canceled and consequently in flow throughout the circle is zero.

  We can also see the same result if we see that within each surface there are no charges, so the right of the Zero Gas Law

From the above described the two surfaces have zero flow

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In an "atom smasher," two particles collide head on at relativistic speeds. If the velocity of the first particle is 0.741c to t
galina1969 [7]

Answer:

W_x = 0.9156\ c

Explanation:

given,

velocity of particle 1 = 0.741 c to left

velocity of second particle = 0.543 c to right

relative velocity between the particle = ?

for the relative velocity calculation we have formula

W_x = \dfrac{|u_x - v_x|}{1-\dfrac{u_xv_x}{c^2}}

u_x = 0.543 c

v_x = - 0.741 c

W_x = \dfrac{0.543 c - (-0.741 c)}{1-\dfrac{(0.543 c)(-0.741 c)}{c^2}}

W_x = \dfrac{0.543 c +0.741 c)}{1+\dfrac{(0.543)(0.741)c^2}{c^2}}

W_x = \dfrac{1.284c}{1+0.402363}

W_x = 0.9156\ c

Relative velocity of the particle is W_x = 0.9156\ c

5 0
2 years ago
There is a 90 kg woman attached at the end of a bungee cord (k = 35 N/m) that is experiencing simple harmonic motion. How long d
Cloud [144]

Answer:

The value is T  =10.1 \ s

Explanation:

From the question we are told that

    The mass of the woman is  m  = 90 \  kg  

    The spring constant of the bungee cord is  k  =  35 \  N/  m

Generally the period of the oscillation (i,e time taken to complete on  cycle ) is mathematically represented as

            T  = 2 \pi *  \sqrt{ \frac{m}{k} }

=>      T  = 2 * 3.142  *  \sqrt{ \frac{90 }{ 35} }

=>      T  =10.1 \ s

3 0
3 years ago
When a rocket is 4 kilometers high, it is moving vertically upward at a speed of 400 kilometers per hour. At that instant, how f
Y_Kistochka [10]

Answer:

The angle of elevation of the rocket is increasing at a rate of 48.780º per second.

Explanation:

Geometrically speaking, the distance between the rocket and the observer (r), measured in kilometers, can be represented by a right triangle:

r = \sqrt{x^{2}+y^{2}} (1)

Where:

x - Horizontal distance between the rocket and the observer, measured in kilometers.

y - Vertical distance between the rocket and the observer, measured in kilometers.

The angle of elevation of the rocket (\theta), measured in sexagesimal degrees, is defined by the following trigonometric relation:

\tan \theta = \frac{y}{x} (2)

If we know that x = 5\,km, then the expression is:

\tan \theta = \frac{y}{5}

And the rate of change of this angle is determined by derivatives:

\sec^{2}\theta \cdot \dot \theta = \frac{1}{5}\cdot \dot y

\frac{\dot \theta}{\cos^{2}\theta} = \frac{\dot y}{5}

\frac{\dot \theta\cdot (25+y^{2})}{25} = \frac{\dot y}{5}

\dot \theta = \frac{5\cdot \dot y}{25+y^{2}}

Where:

\dot \theta - Rate of change of the angle of elevation, measured in sexagesimal degrees.

\dot y - Vertical speed of the rocket, measured in kilometers per hour.

If we know that y = 4\,km and \dot y = 400\,\frac{km}{h}, then the rate of change of the angle of elevation is:

\dot \theta = 48.780\,\frac{\circ}{s}

The angle of elevation of the rocket is increasing at a rate of 48.780º per second.

3 0
3 years ago
Calculate the potential energy of a 20-kg sled at 40 meters
irakobra [83]

Answer: 7840N

Explanation:

Given that

Potential energy = ?

Mass of sled = 20-kg

Distance = 40 meters

Acceleration due to gravity = 9.8m/s^2

Recall that potential energy is the energy possessed by a body at rest

i.e potential energy = mass m x acceleration due to gravity g x distance h

P.E = mgh

P.E = 20kg x 9.8m/s^2 x 40m

P.E = 7840N

Thus, the potential energy of the sled is 7840N

5 0
3 years ago
Read 2 more answers
Consider a 150 turn square loop of wire 17.5 cm on a side that carries a 42 A current in a 1.7 T. a) What is the maximum torque
ankoles [38]

Answer:

(a) 328 Nm

(b) 79.35 Nm

Explanation:

N = =150, side = 17.5 cm = 0.175 m, i = 42 A, B = 1.7 T

A = side^2 = 0.175^2 = 0.030625 m^2

(a) Torque = N x i x A x B x Sinθ

For maximum torque, θ = 90 degree

Torque = 150 x 42 x 0.030625 x 1.7 x Sin 90

Torque = 328 Nm

(b) θ = 14 degree

Torque =  150 x 42 x 0.030625 x 1.7 x Sin 14

Torque = 79.35 Nm

7 0
3 years ago
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