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Reil [10]
2 years ago
14

A hiker walks 9.4 miles at an angle of 60° south of west. Find the west and south components of the walk. Round your answers to

the nearest tenth.
Ax = miles
Physics
1 answer:
Tanzania [10]2 years ago
4 0

Answer:

The west axis component of walk is, A = 5 miles

The east axis component of walk is, B = 8 miles

Explanation:

Given data,

The displacement of hiker south of west, R = 9.4 miles

The angle formed south of west, Ф = 60°

The vector components along,

                   the west axis is

                                      A = R cosФ

                                       A = 9.4 x cos 60

                                       A = 4.7 miles

                                        A = 5 miles

The west axis component of walk is, A = 5 miles

                    the south axis is

                                      B = R sinФ

                                       B = 9.4 sin 60

                                        B = 8.14 miles

                                         B = 8 miles

The east axis component of walk is, B = 8 miles

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iVinArrow [24]

To solve the problem we must know about the relationship between Speed, distance, and Time.

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We know that sped, distance, and time all are in a relationship to each other. this relationship can be given as,

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The speed of the light is 30,060.12 km/sec.

Given to us

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The force exerted by the wind on the sails of a sailboat is Fsail = 330 N north. The water exerts a force of Fkeel = 210 N east.
Elena L [17]

Answer:

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From the question we are told that

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   The force exerted by water is  F_{keel} =  (210  ) \ N \ east

      The mass of the boat(+ crew) is  m_b  =  260  \ kg

Now Force is mathematically represented as

      F =  ma

Now the acceleration towards the north is mathematically represented as

      a_n  =  \frac{F_{sail}}{m_b}

substituting values

       a_n  =  \frac{330 }{260}

      a_n  =  1.269 \ m/s^2

Now the acceleration towards the east is mathematically represented as

       a_e = \frac{F_{keel}}{m_b }

substituting values

      a_e = \frac{210}{260}

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The resultant acceleration is  

      a_r =  \sqrt{a_e^2 + a_n^2}

substituting values

     a_r =  \sqrt{(0.808)^2 + (1.269)^2}

      a_r = 1.50 \ m/s^2

The direction with reference from the north is evaluated as

Apply SOHCAHTOA

        tan \theta =  \frac{a_e}{a_n}

       \theta = tan ^{-1} [\frac{a_e}{a_n } ]

substituting values

     \theta = tan ^{-1} [\frac{0.808}{1.269 } ]

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   \theta =  32.5 6^o

     

   

       

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Answer:

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Explanation:

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Beam length = 3 m

Cable angle = 60° above horizontal

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Torque about the pin = Force × Perpendicular distance of force from pin

Where the force = Force due to gravity or weight, we have

Weight = Mass × Acceleration due to gravity = 1975 × 9.81 = 19374.75 N

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Since the distance from pin to the contact forces between the pin and the beam is 0, the torque which is force multiplied by perpendicular distance is also 0 N·m

c) To find the expression for the tension force, T we find the sum of the moment forces about the pin as follows

Sum of moments about p is given as follows

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Therefore torque due to tension is given by the following expression

Torque, due \ to \ tension =L\cdot Tsin\theta = \frac{M\cdot L\cdot g}{2}

d) Plugging in the values in the torque due to tension equation, we have;

3\times Tsin60 = \frac{1975\times 3\times 9.81}{2} = 29062.125

Therefore, we make the tension force, T the subject of the formula hence

T= \frac{29062.125}{3 \times sin(60)} = 11186.02 N

8 0
3 years ago
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