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Vlad [161]
3 years ago
15

Identical +7.67 μC charges are fixed to adjacent corners of a square. What charge (magnitude and algebraic sign) should be fixed

to one of the empty corners, so that the total potential at the remaining empty corner is 0 V?
Physics
1 answer:
Svetlanka [38]3 years ago
8 0

Answer:

q₃ = - 13.0935 μC

Explanation:

Given

q₁ = q₂ = +7.67 μC

We use the equation

V = Kq/r

We can apply it as follows

V₁ = K*q₁/r₁ = K*q₁/(√2*L)

V₂ = K*q₂/r₂ = K*q₂/L

V₃ = K*q₃/r₃ = K*q₃/L

Then

V₁ + V₂ + V₃ = 0

⇒   (K*q₁/(√2*L)) + (K*q₂/L) + (K*q₃/L) = 0

⇒   (K/L)*((q₁/√2) + q₂ + q₃) = 0

⇒   (q₁/√2) + q₂ + q₃ = 0

Since q₁ = q₂

⇒   (q₁)((1/√2) + 1) + q₃ = 0

⇒   q₃ = - (q₁)((1/√2) + 1) = +7.67 μC*(1.7071)

⇒   q₃ = - 13.0935 μC

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As we know that as per Newton's II law we have

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The displacement of a car is a function of time as follows: x(t)=25+3.0t², with x is in meters. Find the average velocity betwee
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At t₁ = 1.0s, displacement x₁ is:

x(1)=25+3(1)^{2}

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At t₂ = 4.0s:

x(4)=25+3(4)^{2}

x(4) = 73

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What will most likely occur if sulfur forms an ionic bond with another element?
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Read 2 more answers
An airplane is flying 340 km/hr at 12o east of north. the wind is blowing 40 km/hr at 34o south of east. what is the plane's act
seropon [69]
Define an x-y coordinate system such that
The positive x-axis = the eastern direction, with unit vector  \hat{i}.
The positive y-axis = the northern direction, with unit vector \hat{j}.

The airplane flies at 340 km/h at 12° east of north. Its velocity vector is
\vec{v}_{1} = 340(sin(15^{o})\hat{i} + cos(15^{o})\hat{j} ) = 88\hat{i} + 328.4\hat{j}

The wind blows at 40 km/h in the direction 34° south of east. Its velocity vector is
\vec{v}_{2} =40(cos(34^{o})\hat{i} - sin(24^{o})]\hat{j}) = 33.1615\hat{i} -22.3677\hat{j})

The plane's actual velocity is the vector sum of the two velocities. It is
\vec{v}=\vec{v}_{1}+\vec{v}_{2} = 121.1615\hat{i}+306.0473\hat{j}

The magnitude of the actual velocity is
v = √(121.1615² + 306.0473²) = 329.158 km/h

The angle that the velocity makes north of east is
tan⁻¹ (306.04733/121.1615) = 21.6°

Answer:
The actual velocity is 329.2 km/h at 21.6° north of east.
8 0
3 years ago
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