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2 years ago

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Two rigid rods are oriented parallel to each other and to the ground. The rods carry the same current in the same direction. The

length of each rod is 0.85 m, and the mass of each is 0.073 kg. One rod is held in place above the ground, while the other floats beneath it at a distance of 8.2 × 10−3 m. Determine the current in the rods.

1 answer:

bazaltina [42]2 years ago

5 0

Answer:

The current in the rods is 171.26 A.

Explanation:

Given that,

Length of rod = 0.85 m

Mass of rod = 0.073 kg

Distance $d = 8.2\times10^{-3}\ m$

The rods carry the same current in the same direction.

We need to calculate the current

I is the current through each of the wires then the force per unit length on each of them is

Using formula of force

$\dfrac{F}{L}=\dfrac{\mu_{0}I^2}{2\pi d}$

$\dfrac{mg}{L}=\dfrac{\mu_{0}I^2}{2\pi d}$

Where, m = mass of rod

l = length of rod

Put the value into the formula

$I^2=\dfrac{mgd}{\mu L}$

$I^2=\dfrac{0.073\times9.8\times8.2\times10^{-3}}{2\times10^{-7}}$

$I=\sqrt{29331.4}$

$I=171.26\ A$

Hence, The current in the rods is 171.26 A.

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Lost mass is proportional to the square root of the distance covered while lifting:

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Mass of the floor containing bag at a height h:

$m(h) = 60 - k\sqrt{h}$

Work done is given by:

$W = \int_{0}^{h}m(h)gdh$

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$W = 9.8\times ([60\times 9 - 0] + \frac{8}{3}[9^{\frac{3}{2}} - 0^{\frac{3}{2}}])$

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Explanation:

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We need to find the initial velocity $V_{o}$ at which the water ballon was thrown, and we can find it by the following equation:

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Where $T=2t$ is the total time the water ballon is on air

On the other hand, when we talk about parabolic motion (as in this situation) the water ballon reaches its maximum height just in the middle of this parabola, when $V=0$ and the time $t$ is half the time $T$ it takes the complete parabolic path.

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Substituting (4) in (1):

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Answer:

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Making acceleration,a the subject of the formula and substituting the parameters into equation (1);

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