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Ksju [112]
3 years ago
10

If a crow flies west for 60 km and then south for 45 km, what is the direction of its displacement?

Physics
1 answer:
son4ous [18]3 years ago
7 0
That's 105 km that he flew, or 65.2 miles !  I'm absolutely positive
that the crow must have landed and gotten some rest when you
weren't looking.  But that had no effect on his displacement when
he got where he was going, so we can continue to solve the problem:


The displacement is the distance and direction from the place
where the crow took off to the place where he landed.

-- It's distance is the hypotenuse of the right triangle whose legs
are 60 km and 45 km.

        D²  =  (60 km)²  +  (45 km)²

              =    3,600 km²  +  2,025 km²  =  5,625 km²

         D  =  √(5625 km²)  =  75 km .    
 
-- It's direction is the angle whose tangent is  (45 S / 60 W).

         tan⁻¹ (45/60)  =  tan⁻¹ (0.75)  =  36.9° south of west

                                                         =  53.1° west of south.

                                                         =  not exactly southwest but close.
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<h3>Force required to pull one end at a constant speed</h3>

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1.) What statement best explains the type of population growth seen in Belize according to the 2010 age structure diagram?
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Complete Question

The complete question is shown on the first uploaded image

Answer:

The components of reaction at the fixed support are  

    A_{(x)}  = 400  \ N ,  A_{(y)}  = -500  \ N ,  A_{(z)}  = 600  \ N ,  M_x  = 1225 \  N\cdot m , M_y  = 750 \  N\cdot m ,  M_z  = 0 \  N\cdot m

Explanation:

Looking at  the diagram uploaded we see that there are two  forces acting along the x-axis on the fixed support    

   These force are  400 N  and  A_{(x)} [ i.e the reactive force of  400 N  ]

Hence the sum of forces along the x axis is mathematically represented as

        A_{(x)}  - 400  = 0

=>     A_{(x)}  = 400  \ N

Looking at  the diagram uploaded we see that there are two  forces acting along the y-axis on the fixed support  

   These force are  500 N  and  A_{(y)} [ i.e the force acting along the same direction with 500 N   ]

Hence the sum of forces along the x axis is mathematically represented as

        A_{(y)}  + 500  = 0

=>     A_{(y)}  = -500  \ N

Looking at  the diagram uploaded we see that there are two  forces acting along the z-axis on the fixed support  

       These force are  600 N  and  A_{(z)} [ i.e the reactive force of  600 N  ]

Hence the sum of forces along the x axis is mathematically represented as

        A_{(z)}  - 600  = 0

=>     A_{(z)}  = 600  \ N

Generally taking moment about A along the x-axis we have that

    \sum M_x  = M_x  - 500 (0.75 + 0.5) + 600 ( 1 ) = 0

=>   M_x  = 1225 \  N\cdot m

Generally taking moment about A along the y-axis we have that

    \sum M_y  = M_y  - 400 (0.75 ) + 600 ( 0.75 ) = 0

=>   M_y  = 750 \  N\cdot m

Generally taking moment about A along the z-axis we have that

    \sum M_z  = M_z = 0

=>   M_z  = 0 \  N\cdot m

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