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17 June, 04:41

The force exerted by the wind on the sails of a sailboat is 360 N north. The water exerts a force of 170 N east. If the boat (including its crew) has a mass of 310 kg, what are the magnitude and direction of its acceleration?

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  1. 17 June, 04:50
    0
    The magnitude and direction of its acceleration is 1.284 m/s² North-East

    Explanation:

    Given;

    The force exerted by the wind on the sails of a sailboat to be 360 N north

    The force exerted by water on the sails of a sailboat to be 170 N east.

    The resultant force on the boat is can be calculated by Pythagoras theorem, if these two forces makes right angled-triangle.

    R² = N² + E²

    R² = (360) ² + (170) ²

    R² = 158500

    R = √ (158500)

    R = 398.121 N

    If the If the boat (including its crew) has a mass of 310 kg, then the magnitude and direction of its acceleration becomes;

    a = F/m

    a = 398.121/310

    a = 1.284 m/s² North-East
  2. 17 June, 04:58
    0
    The North and East forces are 90° apart, making a right triangle with the resultant as the hypotenuse. Using trigonometry equations,

    (170) ² + (360) ² = F²

    28900 + 129600 = F²

    √ (158500) = F

    398.12 N = F

    Remember,

    F = ma

    398.12 N = (310 kg) * a

    398.12/310kg = a

    1.28 m/s² = a

    The sailboat will be heading North East. The angle of the boats trajectory use inverse tangent function.

    tan (Ф) = opposite/adjacent

    = arctan (opposite/adjacent)

    = arctan (360/170)

    = 64.7° North East.
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