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ad-work [718]
2 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]2 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]2 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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3. A large crane lifts a 25,000 kg mass in the air. The amount of work that must be done by the
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\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}

Actually Welcome to the concept of Efficiency.

Here we can see that, the Input work is given as 2.2 x 10^7 J and the efficiency is given as 22%

The efficiency is => 22% => 22/100.

so we get as,

E = W(output) /W(input)

hence, W(output) = E x W(input)

so we get as,

W(output) = (22/100) x 2.2 x 10^7

=> W(output) = 0.22 x 2.2 x 10^7 => 0.484 x 10^7

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2 years ago
Ricardo and Jane are standing under a tree in the middle of a pasture. An argument ensues, and they walk away in different direc
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Answer:

the direction that should be walked by Ricardo to go directly to Jane is 23.52 m, 24° east of south

Explanation:

given information:

Ricardo walks 27.0 m in a direction 60.0 ∘ west of north, thus

A= 27

Ax =  27 sin 60 = - 23.4

Ay = 27 cos 60 = 13.5

Jane walks 16.0 m in a direction 30.0 ∘ south of west, so

B = 16

Bx = 16 cos 30 = -13.9

By = 16  sin 30 = -8

the direction that should be walked by Ricardo to go directly to Jane

R = √A²+B² - (2ABcos60)

   = √27²+16² - (2(27)(16)(cos 60))

   = 23.52 m

now we can use the sines law to find the angle

tan θ = \frac{R_{y} }{R_{x} }

         = By - Ay/Bx -Ax

         = (-8 - 13.5)/(-13.9 - (-23.4))

     θ  = 90 - (-8 - 13.5)/(-13.9 - (-23.4))

         = 24° east of south

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the arm of a crane is 15.0 m long and makes an angle of 70.0 degrees with the horizontal. Assume that the maximum load for the c
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To find max torque you need component of force that is perpendicular to the direction of vector r. For second part just set the equation above equal to 2308 and solve for Fmax. Only difference is the angle

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