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pashok25 [27]
3 years ago
8

.5(x + 3) ^{2} " alt="x + 5 = 0.5(x + 3) ^{2} " align="absmiddle" class="latex-formula">
Physics
1 answer:
nikitadnepr [17]3 years ago
6 0
Solve for x over the real numbers = complete the square
x + 5 = 0.5 (x + 3)^2

0.5 (x + 3)^2 = 1/2 (x + 3)^2:
x + 5 = 1/2 (x + 3)^2

Expand out terms of the right hand side:
x + 5 = x^2/2 + 3 x + 9/2

Subtract x^2/2 + 3 x + 9/2 from both sides:
-x^2/2 - 2 x + 1/2 = 0

Multiply both sides by -2:
x^2 + 4 x - 1 = 0

Add 1 to both sides:
x^2 + 4 x = 1

Add 4 to both sides:
x^2 + 4 x + 4 = 5

Write the left hand side as a square:
(x + 2)^2 = 5

Take the square root of both sides:
x + 2 = sqrt(5) or x + 2 = -sqrt(5)

Subtract 2 from both sides:
x = sqrt(5) - 2 or x + 2 = -sqrt(5)

Subtract 2 from both sides:
Answer:  x = sqrt(5) - 2 or x = -2 - sqrt(5)
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Two long, parallel wires are attracted to each other by a force per unit length of 350 µN/m. One wire carries a current of 22.5
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Answer

given,

force per unit length = 350 µN/m

current, I = 22.5 A

y = y = 0.420 m

\dfrac{F}{L}= \dfrac{KI_1I_2}{d}

I_2 = \dfrac{F}{L}\dfrac{d}{KI_1}

I_2 = 350\times 10^{-6}\times \dfrac{0.42}{2 \times 10^{-7}\times 22.5}

    I₂ = 32.67 A

distance where the magnetic field is zero

\dfrac{4\pi \times 10^{-7}\times 32.67}{2\pi y_1}=\dfrac{4\pi \times 10^{-7}\times 22.5}{2\pi (0.42-y_1)}

y_1 = 0.248\ m

there the distance at which the magnetic field is zero in the two wire is at 0.248 m.

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