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What of the following is an accurate statement

1 answer:

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B is right. Because of the larger turns on the secondary the voltage on the secondary is larger. This is what is meant by "step-up transformer" The voltage is stepped up to a higher value

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Water flowing through a cylindrical pipe suddenly comes to a section of the pipe where the diameter decreases to 86% of its prev

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The speed in the smaller section is $43.2\,\frac{m}{s}$

Explanation:

Assuming all the parts of the pipe are at the same height, we can use continuity equation for incompressible fluids:

$\Delta Q=0$ (1)

With Q the flux of water that is $Av$ with A the cross section area and v the velocity, so by (1):

$A_{2}v_{2}-A_{1}v_{1}=0$

subscript 2 is for the smaller section and 1 for the larger section, solving for $v_{2}$ :

$v_{2}=\frac{A_{1}v_{1}}{A_{2}}$ (2)

The cross section areas of the pipe are:

$A_{1}=\frac{\pi}{4}d_{1}^{2}$

$A_{2}=\frac{\pi}{4}d_{2}^{2}$

but the problem states that the diameter decreases 86% so $d_{2}=0.86d_{1}$ , using this on (2):

$v_{2}=\frac{\frac{\pi}{4}d_{1}^{2}v_{1}}{\frac{\pi}{4}d_{2}^{2}}=\frac{\cancel{\frac{\pi}{4}d_{1}^{2}}v_{1}}{\cancel{\frac{\pi}{4}}(0.86\cancel{d_{1}})^{2}}\approx1.35v_{1}$

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