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MrRa [10]
2 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]2 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:

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Explanation:

From the question we are told that

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Generally from kinematic equation we have that

      v^2 = u^2 + 2as

Here u is the initial  speed of the aircraft with value 0 m/ s give that the aircraft started from rest

So  

    147^2 = 0^2 + 2* a* 0.208

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A) Determine the x and y-components of the ball's velocity at t = 0.0s, 2.0, 3.0 secs.
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The kinematic relationships we can find the position, acceleration and launch angle of the body on the planet Exidor.

a) the position are

      time (s)  x (m)   y(m)

        0            0          0

        2.0         3.6        1.2

        3.0         5.4        0.9

b) The aceleration is  g = 0.6 m / s²

c) The launch angle      θ = 33.7º

given parameters

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  • the movement times t = 1.0s, 2.0s and 3.0 s

to find

    a) position

    b) acceleration

    c) launch angle

Projectile launch is an application of kinematics to the movement of the body in two dimensions where there is no acceleration on the x axis and the y axis has the planet's gravity acceleration

b) To calculate the acceleration of the plant acting on the y-axis, we use that the vertical velocity of the body at the highest point is zero.

         v_y = v_{oy} - g t

where v and v({oy}  are the velocities of the body, g the acceleration of the planet's gravity and t the time

          0 = v_{oy} - gt

           g = v_{oy} / t

from the graph we observe that the highest point occurs for t = 2.0 s

           g = 1.2 / 2.0

           g = 0.6 m / s²

 

a) The position is requested for several times

X axis

in this axis there is no acceleration so we can use the uniform motion relationships

          vₓ = x / t

          x = vₓ t

where x is the position, vx is the velocity and t is the time

we calculate for the time

t = 0.0 s

          x₀ = 0

           

t = 2.0 s

          x₂ = 1.8 2

          x₂ = 3.6 m

t = 3.0 s

          x₃ = 1.8 3

          x₃ = 5.4 m

Y axis

In this axis there is the acceleration of the planet, let us use for the position the relation

          y = v_{oy} t - ½ g t²

t = 0.0 s

          y₀ = 0

          y₀ = 0 m

t = 2.0 s

         y₂ = 1.2 2 - ½ 0.6 2²

         y₂ = 1.2 m

t = 3.0 s

        y₃ = 1.2  3 - ½  0.6  3²

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        θ = tan⁻¹ \frac{v_y}{v_x}

        θ = tan⁻¹ \frac{1.2}{1.8}

        θ = 33.7º

measured counterclockwise from the positive side of the x-axis

With the kinematic relationships we can find the position, acceleration and launch angle of the body on the planet Exidor.

a) the position are

      time (s)  x (m)   y(m)

        0            0          0

        2.0         3.6        1.2

        3.0         5.4        0.9

b) The aceleration is  g = 0.6 m / s²

c) The launch angle      θ = 33.7ºto)

learn more about projectile launch here:

brainly.com/question/10903823

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