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ad-work [718]
3 years ago
9

Three equal negative point charges are placed at three of the corners of a square of side d. What is the magnitude of the net el

ectric field at the center of the square? (k=1/4πϵ0=8.99×109N⋅m2/C2 )
Express your answer in terms of the variables d, q, and appropriate constants.
Physics
2 answers:
Rina8888 [55]3 years ago
8 0
<span>this  may help you
As far as the field goes, the two charges opposite each other cancel!
So E = kQ / d² = k * Q / (d/√2)² = 2*k*Q / d² ◄
and since k = 8.99e9N·m²/C²,
E = 1.789e10N·m²/C² * Q / d² </span>
Vaselesa [24]3 years ago
7 0

Answer:

E = 2kq /d^2

Explanation:

Let d = side of the square

Distance between the center and corner of the square = a

a = √(d^2 + d^2) / 2

a = √2d^2 / 2

a = (√2 * d) /2

a = d /√2

Let the magnitude of an electric field = E

E = kq/a^2

Put a = d/√2

E = kq / (d/√2)^2

E = kq / d^2 / 2

E = 2kq / d^2

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M A sinusoidal wave on a string is described by the wave function
IceJOKER [234]

The frequency of the wave is determined as 7.96 Hz.

<h3>Frequency of the wave</h3>

The frequency of the wave is calculated as follows;

y = A sin(ωt - kx)

where;

  • A is amplitude of the wave
  • ω is angular speed of the wave

ω = 2πf

f = ω/2π

f = (50)/(2π)

f = 7.96 Hz

Thus, the frequency of the wave is determined as 7.96 Hz.

Learn more about frequency of waves here: brainly.com/question/6297363

#SPJ4

5 0
2 years ago
A glass plate whose index of refraction is 1.66 is immersed in a liquid. The surface of the glass is inclined at an angle of 44º
Anna35 [415]

Answer:

C. 1.15

Explanation:

The formula for the critical angle is:

{sin\theta_{critical}}=\frac {n_r}{n_i}

Where,  

{\theta_{critical}} is the critical angle

n_r is the refractive index of the refractive medium.  ( Alcohol , ? )

n_i is the refractive index of the incident medium.  ( Glass, 1.66 )

<u>Critical angle =  44º</u>

Applying in the formula as:

{sin\theta_{critical}}=\frac {1}{1.523}

{sin\ 44^0}=\frac {n_r}{1.66}

{n_r}=1.15

5 0
3 years ago
HELP DUE TONIGHT<br><br> What is friction? How does friction affect the motion of an object?
hram777 [196]

Answer:

Friction always acts in the direction opposing motion. This means if friction is present, it counteracts and cancels some of the force causing the motion (if the object is being accelerated).

Explanation:

Friction is the force resisting the relative motion of solid surfaces, fluid layers, and material elements sliding against each other.

3 0
4 years ago
Decreasing a telescope's eyepiece focal length will _____.
Novay_Z [31]
I believe the correct answer from the choices listed above is the first option. Decreasing a telescope's eyepiece focal length will increase magnification. <span>The magnification of the </span>telescope<span> image is (</span>focal length<span> of the objective) divided by (</span>focal length <span>of the </span>eyepiece<span>). Hope this answers the question.</span>
7 0
3 years ago
Read 2 more answers
A wave with a frequency of 1200 Hz propagates along a wire that is under a tension of 800 N. Its wavelength is 39.1 cm. What wil
mash [69]

Answer:

The wavelength will be 33.9 cm

Explanation:

Given;

frequency of the wave, F = 1200 Hz

Tension on the wire, T = 800 N

wavelength, λ = 39.1 cm

F = \frac{ \sqrt{\frac{T}{\mu} }}{\lambda}

Where;

F is the frequency of the wave

T is tension on the string

μ is mass per unit length of the string

λ is wavelength

\sqrt{\frac{T}{\mu} } = F \lambda\\\\\frac{T}{\mu} = F^2\lambda^2\\\\\mu =  \frac{T}{F^2\lambda^2} \\\\\frac{T_1}{F^2\lambda _1^2} = \frac{T_2}{F^2\lambda _2^2} \\\\\frac{T_1}{\lambda _1^2} = \frac{T_2}{\lambda _2^2}\\\\T_1 \lambda _2^2 = T_2\lambda _1^2\\\\

when the tension is decreased to 600 N, that is T₂ = 600 N

T_1 \lambda _2^2 = T_2\lambda _1^2\\\\\lambda _2^2  = \frac{T_2\lambda _1^2}{T_1} \\\\\lambda _2 = \sqrt{\frac{T_2\lambda _1^2}{T_1}} \\\\\lambda _2 = \sqrt{\frac{600* 0.391^2}{800}}\\\\\lambda _2  = \sqrt{0.11466} \\\\\lambda _2  =0.339 \ m\\\\\lambda _2  =33.9  \ cm

Therefore, the wavelength will be 33.9 cm

5 0
3 years ago
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