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Katyanochek1 [597]
3 years ago
7

A spring has a spring constant of 256 n/m. how far must it be stretched to give it an elastic potential energy of 48 j?

Physics
1 answer:
pychu [463]3 years ago
4 0
Elastic potential energy=1/2kx^2 (k is constant and x is how far it has to be stretched. So 48=1/2(256)x^2 which is 0.6123
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Answer:

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To know what is the answer you first take into account the magnetic field generated by each current, for a distance of d:

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Next, you use the formula for the magnetic force produced by the wires:

\vec{F_B}=I\vec{L}\ X \vec{B}

if the direction of the L vector is in +k direction, the first wire produced a magnetic field with direction +y, that is, +j and the second wire produced magnetic field with direction -y, that is, -j (this because the direction of the magnetic field is obtained by suing the right hand rule). Hence, the direction of the magnetic force on each wire, produced by the other one is:

\vec{F_{B1}}=I_1L\hat{k}\ X\ B_2(-\hat{j})=I_1LB_2\hat{i}=(2A^2)\frac{L\mu_o}{2\pi d}\hat{i}\\\\\vec{F_{B2}}=I_2L\hat{k}\ X\ B_2(\hat{j})=I_2LB_1\hat{i}=-(2A^2)\frac{L\mu_o}{2\pi d}\hat{i}

Hence, due to this result you have that:

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lisabon 2012 [21]
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