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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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Vsevolod [243]
D=M/V = 0.7/2 = 0.35
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3 years ago
A 3-kg skateboard is rolling down the sidewalk at 4 m/s when it collides with a 1-kg skateboard that was initially at rest. If t
irga5000 [103]

Answer:

2 m/s

Explanation:

Parameters given:

Mass of first skateboard, m = 3 kg

Initial speed of first skateboard, u = 4 m/s

Mass of second skateboard, M = 1 kg

Initial speed of second skateboard, U = 0 m/s

Final speed of second skateboard, V = 6 m/s

Using the principle of the conservaton of momentum, the total initial momentum is equal to the total final momentum.

Momentum is the product of mass and velocity. This implies that:

m*u + M*U = m*v + M*V

(3*4) + (1*0) = (3*v) + (1*6)

12 + 0 = 3v + 6

=> 3v = 12 - 6

3v = 6

v = 6/3 = 2 m/s

The final speed of the 3 kg skateboard is 2 m/s

8 0
4 years ago
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In a series circuit, an ammeter shows the current leaving the positive end of a cell is 0.5A. What current flows into the negati
Evgesh-ka [11]

Answer:

0.5A as in a series circuit the current is the same everywhere

Explanation:

5 0
3 years ago
The perihelion of the comet TOTAS is 1.69 AU and the aphelion is 4.40 AU. Given that its speed at perihelion is 28 km/s, what is
dybincka [34]

Answer:

The speed at the aphelion is 10.75 km/s.

Explanation:

The angular momentum is defined as:

L = mrv (1)

Since there is no torque acting on the system, it can be expressed in the following way:

t = \frac{\Delta L}{\Delta t}

t \Delta t = \Delta L

\Delta L = 0

L_{a} - L_{p} = 0

L_{a} = L_{p}   (2)

Replacing equation 1 in equation 2 it is gotten:

mr_{a}v_{a} =mr_{p}v_{p} (3)

Where m is the mass of the comet, r_{a} is the orbital radius at the aphelion, v_{a} is the speed at the aphelion, r_{p} is the orbital radius at the perihelion and v_{p} is the speed at the perihelion.          

From equation 3 v_{a} will be isolated:    

v_{a} = \frac{mr_{p}v_{p}}{mr_{a}}

v_{a} = \frac{r_{p}v_{p}}{r_{a}}   (4)    

Before replacing all the values in equation 4 it is necessary to express the orbital radius for the perihelion and the aphelion from AU (astronomical units) to meters, and then from meters to kilometers:

r_{p} = 1.69 AU x \frac{1.496x10^{11} m}{1 AU} ⇒ 2.528x10^{11} m

r_{p} = 2.528x10^{11} m x \frac{1km}{1000m} ⇒ 252800000 km

r_{a} = 4.40 AU x \frac{1.496x10^{11} m}{1 AU} ⇒ 6.582x10^{11} m

r_{p} = 6.582x10^{11} m x \frac{1km}{1000m} ⇒ 658200000 km  

     

Then, finally equation 4 can be used:

v_{a} = \frac{(252800000 km)(28 km/s)}{(658200000 km)}

v_{a} = 10.75 km/s

Hence, the speed at the aphelion is 10.75 km/s.

       

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Answer:    yeah it is reflection

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