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MrRa [10]
3 years ago
6

A boat race runs along a triangular course marked by buoys A, B, and C. The race starts with the boats headed west for 3700 mete

rs. The other two sides of the course lie to the north of the first side, and their lengths are 1700 meters and 2900 meters. Draw a figure that gives a visual representation of the situation. Then find the bearings for the last two legs of the race.

Physics
1 answer:
ale4655 [162]3 years ago
7 0

Answer:

The  last two bearings are

49.50° and 104.02°

Explanation:

Applying the Law of cosine (refer to the figure attached):

we have

x² = y² + z² - 2yz × cosX

here,

x, y and z represents the lengths of sides opposite to the angels X,Y and Z.

Thus we have,

cos X=\frac{x^2-y^2-z^2}{-2yz}

or

cos X=\frac{y^2 + z^2-x^2}{2yz}

substituting the values in the equation we get,

cos X=\frac{2900^2 + 3700^2-1700^2}{2\times 2900\times 3700}

or

cos X=0.8951

or

X = 26.47°

similarly,

cos Y=\frac{1700^2 + 3700^2-2900^2}{2\times 1700\times 3700}

or

cos Y=0.649

or

Y = 49.50°

Consequently, the angel Z = 180° - 49.50 - 26.47 = 104.02°

The bearing of 2 last legs of race are angels Y and Z.

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Answer:

You're strong.

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Fan object moves in uniform circular motion in a circle of radius R=200 meters, and the objectes 5.00 seconds to
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Answer:

The centripetal acceleration of the object is 31550.72\ m/s^2.  

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\omega=\dfrac{\Delta \theta}{\Delta t}\\\\\omega=\dfrac{2\pi \times 10}{5}\\\\\omega=12.56\ rad/s

The centripetal acceleration of the object is given by :

a=\omega^2 r\\\\a=(12.56)^2 \times 200\\\\a=31550.72\ m/s^2

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4. How much force is required to stop a 60 kg person traveling at 30 m/s during a time of a)
11111nata11111 [884]

Explanation:

F = ma, and a = Δv / Δt.

F = m Δv / Δt

Given: m = 60 kg and Δv = -30 m/s.

a) Δt = 5.0 s

F = (60 kg) (-30 m/s) / (5.0 s)

F = -360 N

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F = (60 kg) (-30 m/s) / (0.50 s)

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3 years ago
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