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Llana [10]
2 years ago
15

For what value of the ratio r/a of plate radius to separation between the plates does the electric field at the point x=a/2 on t

he x axis differ by 1 percent from the result η/ϵ0 for infinite sheets?
Physics
1 answer:
SIZIF [17.4K]2 years ago
6 0

solution:

E\delta =\frac{R}{\epsilon0}(1-\frac{A}{\sqrt{4R^{2}}-ac}

=\frac{R}{\epsilon0}(1-\frac{1}{\sqrt{4r^{2}/^{_a{2}}+1}})

=\frac{R}{\epsilon0}(1-\frac{1}{\sqrt{4x^2+1}})

x=\frac{r}{a}

infinite case,

Ei=\frac{r}{\epsilon0}

\therefore e\delta =ei(1-\frac{1}{\sqrt{4x^{2}+1}})

we have to find x when,

ei-e\delta =1% ,y=ei=1/100 ei

or,ei-ei+\frac{ei}{\sqrt{4x^2+1}} = 1/100ei

\frac{1}{\sqrt{4x^2+1}}=\frac{1}{100}

4x^2+1 =10^4

x=\frac{\sqrt{\frac{10^4-1}{4}}}=49.99\approx 50

\therefore \frac{r}{a}\approx 50

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A 900-kg car traveling east at 15.0 m/s collides with a 750-kg car traveling north at 20.0 m/s. The cars stick together. Assume
dalvyx [7]

Explanation:

It is given that,

Mass of the car 1, m_1=900\ kg

Initial speed of car 1, u_1=15i\ m/s (east)

Mass of the car 2, m_2=750\ kg

Initial speed of car 2, u_1=20j\ m/s (north)

(b) As the cars stick together. It is a case of inelastic collision. Let V is the common speed after the collision. Using the conservation of momentum as :

m_1u_1+m_2u_2=(m_1+m_2)V

900\times 15i +750\times 20j=(900+750)V

13500i+15000j=1650V

V=(8.18i+9.09j)\ m/s

The magnitude of speed,

|V|=\sqrt{8.18^2+9.09^2}

V = 12.22 m/s

(b) Let \theta is the direction the wreckage move just after the collision. It is given by :

tan\theta=\dfrac{v_y}{v_x}

tan\theta=\dfrac{9.09}{8.18}

\theta=48.01^{\circ}

Hence, this is the required solution.

4 0
2 years ago
luggage handler pulls a 20.0 kg suitcase up a ramp inclined at 34.0 ∘ above the horizontal by a force F⃗ of magnitude 165 N that
Andreas93 [3]

Answer:

a)  W = 643.5 J, b) W = -427.4 J  

Explanation:

a) Work is defined by

       W = F. x = F x cos θ

in this case they ask us for the work done by the external force F = 165 N parallel to the ramp, therefore the angle between this force and the displacement is zero

       W = F x

let's calculate

       W = 165  3.9

        W = 643.5 J

b) the work of the gravitational force, which is the weight of the body, in ramp problems the coordinate system is one axis parallel to the plane and the other perpendicular, let's use trigonometry to decompose the weight in these two axes

       sin θ = Wₓ / W

       cos θ = Wy / W

        Wₓ = W sinθ = mg sin θ

        Wy = W cos θ

the work carried out by each of these components is even Wₓ, it has to be antiparallel to the displacement, so the angle is zero

      W = Wₓ x cos 180

      W = - mg sin 34  x

     

let's calculate

       W = -20 9.8 sin 34 3.9

        W = -427.4 J

The work done by the component perpendicular to the plane is ero because the angle between the displacement and the weight component is 90º, so the cosine is zero.

3 0
2 years ago
Suppose you wanted to hold up an electron against the force of gravity by the attraction of a fixed proton some distance above i
SCORPION-xisa [38]

Answer:

The value is  r =  5.077 \  m

Explanation:

From the question we are told that

   The  Coulomb constant is  k =  9.0 *10^{9} \  N\cdot  m^2  /C^2

   The  charge on the electron/proton  is  e =  1.6*10^{-19} \  C

    The  mass of proton m_{proton} =  1.67*10^{-27} \  kg

    The  mass of  electron is  m_{electron } =  9.11 *10^{-31} \ kg

Generally for the electron to be held up by the force gravity

   Then    

       Electric force on the electron  =  The  gravitational Force

i.e  

            m_{electron} *  g  = \frac{ k *  e^2  }{r^2 }

         \frac{9*10^9 *  (1.60 *10^{-19})^2  }{r^2 }  =     9.11 *10^{-31 }  *  9.81

         r =  \sqrt{25.78}

         r =  5.077  \  m

7 0
2 years ago
Please re-explain the following phrases in terms of momentum
zheka24 [161]
A. An object in motion will remain in motion unless acted upon by an outside force : The momentum of an object is constant unless an outside force acts on the object.

B. Force is defined as mass times acceleration : the rate of change of the momentum of a particle is proportional to the force F acting on it, hence the force is equal to <span>mass times acceleration.

C. </span>For every action there is an equal and opposite reaction : <span>to every action force there is an equal and opposite reaction force. </span>
8 0
2 years ago
The pupil of the human eye can vary in diameter from 2.00 mm in bright light to 8.00 mm in dim light. The eye has a focal length
Zarrin [17]

The indicated data are of clear understanding for the development of Airy's theory. In optics this phenomenon is described as an optical phenomenon in which The Light, due to its undulatory nature, tends to diffract when it passes through a circular opening.

The formula used for the radius of the Airy disk is given by,

y_r=1.22\frac{\lambda f}{d}

Where,

y_r = Range of the radius

\lambda = wavelength

f= focal length

Our values are given by,

State 1:

d=2.00mm = 2*10^{-3}m

f= 25mm = 25*10^{-3}m

\lambda = 750nm = 750*10^{-9}m

State 2:

d=8.00mm = 8*10^{-3}m

f= 25mm = 25*10^{-3}m

\lambda = 390nm = 390*10^{-9}

Replacing in the first equation we have:

y_{r1} = 1.22\frac{(750*10^{-9})(25*10^{-3})}{2*10^{-3}}

y_{r1}= 11.4\mu m

And also for,

y_{r2} =1.22\frac{(390*10^{-9})(25*10^{-3})}{8*10^{-3}}

y_{r2} = 1.49\mu m

Therefor, the airy disk radius ranges from 1.49\mu m to 11.4\mu m

7 0
2 years ago
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