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skelet666 [1.2K]
3 years ago
15

John has a boat that will travel at the rate of 15 kph in still water. he can go upstream for 35 km in the same time it takes to

go 140 km downstream. how fast is his boat traveling when he goes downstream? 22 kph 24 kph 26 kph
Physics
1 answer:
MA_775_DIABLO [31]3 years ago
5 0
Let F = the downstream speed of the water. 

<span>Then the boat's upstream speed is: 15 - F </span>
<span>The boat's downstream speed is: 15 + F </span>


<span>Assume both the journeys mentioned take T hours, then using "speed x time = distance" we get: </span>

<span>Downstream journey: (15 + F)T = 140 </span>
<span>Upstream journey: (15 - F)T = 35 </span>


<span>Add the two formulae together: </span>

<span>(15 + F)T + (15 - F)T = 140 + 35 </span>

<span>15T + FT + 15T - FT = 175 </span>

<span>30T = 175 </span>

<span>T = 35/6 </span>


<span>Use one of the equations to find F: </span>

<span>(15 + F)T = 140 </span>

<span>15 + F = 140/T </span>
<span>F = 140/T - 15 </span>
<span>F = 140/(35/6) - 15 </span>
<span>F = 24 - 15 </span>
<span>F = 9 </span>

<span>i.e. the downstream speed of the water is 9 kph </span>

<span>Therefore, the boat's speed downstream is 15 + F = 15 + 9 = 24 kph.
the answer is:                       *24kph*</span>
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If you assume that the magnetic field is measured in the up-right corner of the square, the wire to the left generates a magnetic field (in the corner in which you measure B) with direction upward (+ j), the wire down (down-right) generates a magnetic field with direction to the left (- i)  and the third wire generates a magnetic field with a direction that is 45° over the horizontal in the left direction (you can notice that in the image attached below). The total magnetic field will be:

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B_T=\frac{\mu_o I}{2\pi}[(-\frac{1}{r_2}-\frac{cos45}{r_3})\hat{i}+(\frac{1}{r_1}+\frac{sin45}{r_3})\hat{j}}]\\\\B_T=\frac{(4\pi*10^{-7}T/A)(6.0A)}{2\pi}[(-\frac{1}{0.09m}-\frac{cos45}{0.127m})\hat{i}+(\frac{1}{0.09m}+\frac{sin45}{0.127m})]\\\\B_T=\frac{(4\pi*10^{-7}T/A)(6.0A)}{2\pi}[-16.67\hat{i}+16.67\hat{j}]\\\\B_T=2.0*10^-5[-\hat{i}+\hat{j}]T

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