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2 years ago

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You are stationary on the ground and are watching a bird fly horizontally towards you at a rate of 2 m/s. The bird is located 20

m above your head. How fast does the angle of elevation of your head change when the horizontal distance between you and the bird is 110 m

1 answer:

Ira Lisetskai [31]2 years ago

3 0

The angle of elevation changes by 0.0032 rad / sec

Let x (t) be the horizontal distance from you to bird.

Let θ be the angle of elevation of your head.

Then .tan θ = 20 /x

Differentiating

Sec² = dθ / dt = -20/x².dx/dt

dθ / dt = -20/x².dx/dt.cos²θ

dx/dt = -2m/s x = 110m

cos²θ = 110²/ (110²) + (20²) = 12100/12500

dθ / dt = 20/ (110)² x (-2) . 12100/12500

= 20 x 2/12500 = 0.0032 radian/s.

Hence ,

The angle of elevation changes by 0.0032 rad / sec

Learn more about angle of elevation:brainly.com/question/16716174

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Answer:

<h2><em>6000 counts per second</em></h2>

Explanation:

If a sample emits 2000 counts per second when the detector is 1 meter from the sample, then;

2000 counts per second = 1 meter ... 1

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x count per second = 3 meter ... 2

Solving the two expressions simultaneously for x we will have;

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<em>This shows that 6000 counts per second would be observed when the detector is 3 meters from the sample</em>

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3 years ago

1. A pendulum has a period of 3 seconds. What's its frequency? 2. A pendulum has a frequency of 0.25 Hz. What's its period? 3. A

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Answer:

(1) 0.333 Hz

(2) 4 sec

(3) 2 sec, 0.5 Hz

Explanation:

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(2) Frequency of the pendulum is given f = 0.25 Hz

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3 years ago

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2 years ago

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Answer:

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