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snow_tiger [21]
3 years ago
11

It is common to see birds of prey rising upward on thermals. The paths they take may be spiral-like. You can model the spiral mo

tion as uniform circular motion combined with a constant upward velocity. Assume a bird completes a circle of radius 6.00 m every 5.00 s and rises vertically at a rate of 3.00 m/s.
Physics
1 answer:
TiliK225 [7]3 years ago
6 0

Answer:

a) 8.115 m/s

b) 9.472 m/s^{2}

c) 21.7\°

Explanation:

The rest of the question is written below:

Determine (a) the bird’s speed relative to the ground; (b) the bird’s acceleration (magnitude and direction); and (c) the angle between the bird’s velocity vector and the horizontal.

<u>And we have the following data:</u>

r=6 m  is the radius of the circular path in the x-axis

T=5 s the period of the circular motion of the bird's path

V_{y}=3 m/s the vertical component of the bird's velocity, which is directed upward and is constant.

Now let's begin with the answers:

<h3>a) Bird’s speed relative to the ground</h3>

In order to find this speed, we have to calculate the magnitude of the bird's velocity vector:

V=sqrt{{V_{x}}^{2} + {V_{y}}^{2} (1)

We already know the value of V_{y}. So, we have to find V_{x}.

Since the bird is describing a circular path in the x-axis, V_{x} will be its <u>tangential velocity</u>:

V_{x}=\omega r=\frac{2 \pi}{T} r (2)

Where \omega=\frac{2 \pi}{T} is the <u>birds angular velocity</u>

V_{x}=\frac{2 \pi}{5 s} 6 m (3)

V_{x}=7.539 m/s (4)

Substituting (4) in (1):

V=sqrt{(7.539 m/s)}^{2} + (3 m/s)^{2} (5)

V=8.1147 m/s \approx 8.115 m/s (6) This is the bird's speed relative to the ground

<h3>b) Bird’s acceleration (magnitude and direction)</h3>

Since the vertical component of the bird's velocity is constant, the vertical component of its acceleration is zero:

a_{y}=0 m/s^{2}

However, the bird has radial acceleration a_{r}=a_{x} that results from its rotation on the circular path horizontally:

a_{x}=\frac{{V_{x}}^{2}}{r} (7)

a_{x}=\frac{(7.539 m/s)^{2}}{6 m} (8)

a_{x}=9.472 m/s^{2} (9) This is the magnitude of the bird's acceleration, which is directed to the center of the circular path the bird describes while it is moving upwards in the spiral.

<h3>c) Angle between the bird’s velocity vector and the horizontal</h3>

In order to find the direction of the bird's velocity vector with the horizontal, we have to find the angle between the horizontal and the vertical component of this velocity:

\theta=tan^{-1}(\frac{V_{y}}{V_{x}}) (10)

\theta=tan^{-1}(\frac{3 m/s}{7.539 m/s}) (11)

Finally:

\theta=21.69\° \approx 21.7 \° (12)

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Q = \frac{2\pi L(t_{hf} - t_{cf}) }{\frac{1}{h_{hf}r_1}+\frac{ln(r_2/r_1)}{k_A} + \frac{ln(r_3/r_2)}{k_B} +\frac{1}{h_{cf}r_3}}

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