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Harlamova29_29 [7]

3 years ago

8

Consider n equal positively charged particles each of magnitude Q/n placed symmetrically around a circle of radius a. Calculate

the magnitude of the electric field at a point a distance x from the center of the circle and on the line passing through the center and perpendicular to the plane of the circle. (Use any variable or symbol stated above along with the following as necessary: ke.)

1 answer:

Sonja [21]3 years ago

3 0

Answer: Magnitude of electric field =p × Ke × Q/d^3

Explanation: Using Coulombs law of point charge,each charge on the circle would exert a fieldEc at point given by:

Ec= Ke × (Q/n)/d^2

Where Ke= Coulomb's constant

d= distance between the charges and the point of measurement, P with d^2=a^2+p^2

(Q/n)= Magnitude of the charge.

For charges in a circle,all the force components and direction(x,z) are cancelled by the symmetry,leaving only the vertical force(y-direction)

The Resultant vector will be

Ecy=Eq × sin(theta)

Ecy=Ke × (Q/n)/d^2 ×(p/d)

Adding the forces from all the charges,the magnitude of electric field Ey=n×Ecy

/Ey/= n ×[Ke ×(Q/n)/d^2 ×(p/d)]

/Ey/= p × Ke × (Q/d^3)

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Anaerobic energy

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7 0

3 years ago

A car drives around a curve with radius 539 m at a speed of 32.0 m/s. The road is banked at 5.00°. The mass of the car is 1.40 ×

HACTEHA [7]

Answer:

$f_r = 150.47 N$

Explanation:

given,

r = 539 m

v = 32 m/s

road banked at = 5°

∑ F_x

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∑ F_y = 0

$0 = N cos \theta - f_r sin \theta - mg$

$N = \dfrac{f_rsin \theta + mg}{cos \theta}$

$\dfrac{mv^2}{r}= (\dfrac{f_rsin \theta + mg}{cos \theta})sin \theta + f_r cos \theta$

= $f_r sin \theta tan \theta + mg tan \theta + f_r cos \theta$

$f_r = \dfrac{\dfrac{mv^2}{r}- mg tan\theta}{sin\theta tan \theta + cos \theta}$

$f_r = \dfrac{\dfrac{1.4\times 10^3 \times 32^2}{539}- 1.4\times 10^{3}\times 9.8 \times 0.087}{0.087 \times 0.087 + 0.996}$

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8 0

3 years ago

PLZ HELP AND HURRY

Rainbow [258]

Earth-like planet in another solar system
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3 years ago

After a laser bean passes through two thin parallel slits, thefirst completely dark fringes occur at ± 15.00with the original di

PSYCHO15rus [73]

Answer:

143 °

Explanation:

a ) If d be the distance between slits , λ be wavelength of light used and at angle θ nth dark fringe is formed then

d sinθ = ( 2n+1) λ/2

for first dark fringe

d sinθ = λ/2

d /λ = 1/ 2 sinθ

1 / 2 sin15

= 1.93

b )

For intensity of fringe at angle θ, the relation is

I = I₀ cos²θ

I / I₀ = cos²θ/2

Given I / I₀ =0. 1

0.1 = cos²θ/2

θ/2 = 71.5

θ = 143 °

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3 years ago

Complete the following sentence: The term coherence relates to the phase relationship between two waves. the polarization state

77julia77 [94]

Answer:

<em>the phase relationship between two waves.</em>

<em></em>

Explanation:

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3 years ago