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Talja [164]
3 years ago
9

A man wishes to travel due north in order to cross a river 5 kilometers wide flowing due East at 3 kilometers per hour . if he c

an row at 10 kilometers per hour In still water, find either by scale drawing or calculation
1. the direction in which he must heads his boat in order to get to his destination directly opposite his starting point​
Physics
1 answer:
Liono4ka [1.6K]3 years ago
7 0

Answer:

17.46° W of N

Explanation:

His prow must point westward so that the westward component of his total velocity is equal to the eastward river speed.

3 = 10sinθ

sinθ = 3/10

θ = 17.4576... or about 17.46° W of N

it will take him 5 / 10cos17.46 = 0.52414... hrs. or about 31 min 27 s to cross.

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We are asked to determine the velocity of a rain drop if it falls from 4 km.

To do that we will use the following formula:

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Part B. we are asked to determine the velocity if there is air drag. To do that we will use the following formula:

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\begin{gathered} F_d=drag\text{ force} \\ C=\text{ constant} \\ \rho_{air}=\text{ density of air} \\ A=\text{ area} \\ v=\text{ velocity} \end{gathered}

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Since the diameter of the raindrop is 3 millimeters, the radius is 1.5 mm or 0.0015 meters. Substituting we get:

m=(0.98\times10^3\frac{kg}{m^3})(\frac{4}{3}\pi(0.0015m)^3)

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m=1.39\times10^{-5}kg

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F_d=(1.39\times10^{-5}kg)(9.8\frac{m}{s^2})

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