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mojhsa [17]
3 years ago
11

A line of charge starts at x = +x0 and extends to positive infinity. The linear charge density is λ = λ0x0/x. Determine the elec

tric field at the origin. (Use ke, λ0, and x0 as necessary.)
Physics
2 answers:
kari74 [83]3 years ago
5 0

Explanation:

it is given that, the linear charge density of a charge, \lambda=\dfrac{\lambda_ox_o}{x}

Firstly, we can define the electric field for a small element and then integrate for the whole. The very small electric field is given by :

dE=\dfrac{k\ dq}{x^2}..........(1)

The linear charge density is given by :

\lambda=\dfrac{dq}{dx}

dq=\lambda.dx=\dfrac{\lambda_ox_o}{x}dx

Integrating equation (1) from x = x₀ to x = infinity

E=\int\limits^\infty_{x_o} {\dfrac{k\lambda_ox_o}{x^3}}.dx

E=-\dfrac{k\lambda_ox_o}{2}\dfrac{1}{x^2}|_{x_o}^\infty}

E=\dfrac{k\lambda_o}{2x_o}

Hence, this is the required solution.

Anna007 [38]3 years ago
4 0

Answer:

E=-K_e\dfrac{\lambda_0 }{2x_0}N/C

Explanation:

We know that electric field given as

E=K_e\int_{x_1}^{x_2}\dfrac{\lambda }{x^2}dx

Here given that

\lambda =\dfrac{\lambda _0x_0}{x}

So

E=K_e\int_{x_1}^{x_2}\dfrac{\lambda_0x_0 }{x^3}dx

E=K_e\int_{x_0}^{\infty }\dfrac{\lambda_0x_0 }{x^3}dx

Now by integrating we get

E=-K_e\lambda_0\left [ \dfrac{1 }{2x^2} \right ]^{\infty }_x_0dx

E=-K_e\dfrac{\lambda_0 }{2x_0}dx

So the electric field at the origin E

E=-K_e\dfrac{\lambda_0 }{2x_0}N/C

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A person jumps out a fourth-story window 14 m above a firefighter safety net. The survivor stretches the net 1.8 m before coming
Monica [59]

Answer:

The deceleration is  a =  - 76.27 m/s^2

Explanation:

From the question we are told that

   The height above  firefighter safety net is H  = 14 \ m

   The length by which the net is stretched is s =  1.8 \ m

   

From the law of energy conservation

    KE_T + PE_T =  KE_B + PE_B

 Where KE_T is the kinetic energy of the person before jumping which equal to zero(because to kinetic energy at maximum height )

   and  PE_T is the potential energy of the before jumping  which is mathematically represented at

          PE_T  = mg H

and  KE_B is the kinetic energy of the person just before landing on the safety net  which is mathematically represented at

        KE_B = \frac{1}{2} m v^2

and  PE_B is the potential energy of the person as he lands on the safety net which has a value of zero (because it is converted to kinetic energy )

   So the above equation becomes

          mgH =  \frac{1}{2} m v^2

=>           v =  \sqrt{2 gH }

    substituting values

                v =  16.57 m/s

Applying the equation o motion

             v_f =  v  + 2 a s

Now the final velocity is zero because the person comes to rest

      So

         0 = 16.57 + 2 * a * 1.8

            a =  - \frac{16.57^2 }{2 * 1.8}

            a =  - 76.27 m/s^2

         

         

4 0
3 years ago
What is the speed of a wave that has a wavelength of 225 m and a frequency of 117 hz ? Show your work
avanturin [10]

Answer:

26325 m\s

Explanation:

Data:

v = ?

f = 117 Hz

w = 225

Formula:

v = fw

Solution:

v = ( 117)(225)

v = 26325 m\s <em>A</em><em>n</em><em>s</em><em> </em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em>

6 0
3 years ago
Monochromatic light with a wavelength of 384 nm passes through a single slit and falls on a screen 86 cm away. If the distance o
valkas [14]
This is a Fraunhofer single slit experiment, where the light passing through the slit produces an interference pattern on the screen, and where the dark bands (minima of diffraction) are located at a distance of
y= \frac{m\lambda D}{a}
from the center of the pattern. In the formula, m is the order of the minimum, \lambda the wavelenght, D the distance of the screen from the slit and a the width of the slit.

In our problem, the distance of the first-order band (m=1) is y=0.22 cm. The distance of the screen is D=86 cm while the wavelength is \lambda = 384 nm=384 \cdot 10^{-7}cm. Using these data and re-arranging the formula, we can find a, the width of the slit:
a= \frac{m \lambda D}{y}= \frac{1 \cdot 384 \cdot 10^{-7}cm \cdot 86 cm}{0.22 cm}=0.015 cm
3 0
3 years ago
Does the force of kinetic friction depend on the weight of the block? explain.
drek231 [11]
Yes for an object moving on a horizontal plane, R = mg (where mg = weight). therefore, for an object moving on a horizontal plane: F = μmg



7 0
3 years ago
A homing pigeon starts from rest and accelerates uniformly at +4.00 m/s squared for 10.0 seconds. What is its velocity after the
zavuch27 [327]
'  +4 m/s² ' means that the pigeon's speed is  4 m/s greater every second.

Starting from zero speed, after 10 seconds, its speed is 

                           (10 x 4m/s)  =  40 m/s.

We can't say anything about its velocity, because we have
no information regarding the direction of its flight.
5 0
3 years ago
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