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mixer [17]
2 years ago
15

Two parallel circular rings of radius R have their centres in the X axis separated by a distance L. If each ring carries a unifo

rmly distributed charge Q,find the electric field at points along the X axis
Physics
1 answer:
tatiyna2 years ago
5 0

Answer:

E" =  Q/4πε₀√[(x² + R²)]³(x - (L - x)/√[(L - 2x)L/(x² + R²) + 1]³})

Explanation:

The electric field due to a charged ring of radius R at a distance x from the center of the ring when the axis of the ring is located on the x - axis is

E = Qx/4πε₀[√(x² + R²)]³

Since the rings are separated by a distance L, the electric field at point x due to the second ring is E' = -Q(L - x)/4πε₀[√((L - x)² + R²)]³. It is negative since it points in the negative x - direction.

So, the resultant electric field at x is E" = E + E' = Qx/4πε₀[√(x² + R²)]³ + {-Q(L - x)/4πε₀[√((L - x)² + R²)]³}

E" =  Qx/4πε₀√[(x² + R²)]³ - Q(L - x)/4πε₀√[((L - x)² + R²)]³

E" =  Q/4πε₀(x/√[(x² + R²)]³ - (L - x)/√[((L - x)² + R²)]³})

E" =  Q/4πε₀(x/√[(x² + R²)]³ - (L - x)/√[(L² - 2Lx + x² + R²)]³})

E" =  Q/4πε₀(x/√[(x² + R²)]³ - (L - x)/√[(L - 2x)L + (x² + R²)]³})

E" =  Q/4πε₀√[(x² + R²)]³(x - (L - x)/√[(L - 2x)L/(x² + R²) + 1]³})

So, the electric field at points along the x axis is

E" =  Q/4πε₀√[(x² + R²)]³(x - {(L - x)/√[(L - 2x)L/(x² + R²) + 1]³})

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Throwing a football is an example of force because
siniylev [52]

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i found this provided by the San Francisco 49ers (found on Khan Academy)

Explanation:

I HOPE THIS HELPS!!!

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A small truck has a mass of 2145 kg. How much work is required to decrease the speed of the vehicle from 25.0 m/s to 12.0 m/s on
MAXImum [283]

Answer:

The work required is -515,872.5 J

Explanation:

Work is defined in physics as the force that is applied to a body to move it from one point to another.

The total work W done on an object to move from one position A to another B is equal to the change in the kinetic energy of the object. That is, work is also defined as the change in the kinetic energy of an object.

Kinetic energy (Ec) depends on the mass and speed of the body. This energy is calculated by the expression:

Ec=\frac{1}{2} *m*v^{2}

where kinetic energy is measured in Joules (J), mass in kilograms (kg), and velocity in meters per second (m/s).

The work (W) of this force is equal to the difference between the final value and the initial value of the kinetic energy of the particle:

W=\frac{1}{2} *m*v2^{2}-\frac{1}{2} *m*v1^{2}

W=\frac{1}{2} *m*(v2^{2}-v1^{2})

In this case:

  • W=?
  • m= 2,145 kg
  • v2= 12 \frac{m}{s}
  • v1= 25 \frac{m}{s}

Replacing:

W=\frac{1}{2} *2145 kg*((12\frac{m}{s} )^{2}-(25\frac{m}{s} )^{2})

W= -515,872.5 J

<u><em>The work required is -515,872.5 J</em></u>

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Time dilation: A missile moves with speed 6.5-10 m/s with respect to an observer on the ground. How long will it take the missil
tatyana61 [14]

Answer:

The time taken by missile's clock is 4.6\times 10^{6} s

Solution:

As per the question:

Speed of the missile, v_{m = 6.5\times 10^{3}} m/s

Now,

If 'T' be the time of the frame at rest then the dilated time as per the question is given as:

T' = T + 1

Now, using the time dilation eqn:

T' = \frac{T}{\sqrt{1 + (\frac{v_{m}}{c})^{2}}}

\frac{T'}{T} = \frac{1}{\sqrt{1 + (\frac{v_{m}}{c})^{2}}}

\frac{T + 1}{T} = \frac{1}{\sqrt{1 + (\frac{v_{m}}{c})^{2}}}

1 + \frac{1}{T} = \frac{1}{\sqrt{1 + (\frac{v_{m}}{c})^{2}}}

1 + \frac{1}{T} = (1 + (\frac{v_{m}}{c})^{2})^{- \frac{1}{2}}         (1)

Using binomial theorem in the above eqn:

We know that:

(1 + x)^{a} = 1 + ax

Thus eqn (1) becomes:

1 + \frac{1}{T} = 1 - \frac{- 1}{2}.\frac{v_{m}^{2}}{c^{2}}

T = \frac{2c^{2}}{v_{m}^{2}}

Now, putting appropriate values in the above eqn:

T = \frac{2(3\times 10^{8})^{2}}{(6.5\times 10^{3})^{2}}

T = 4.6\times 10^{6} s

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