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Alex17521 [72]
3 years ago
15

A plane flies horizontally at an altitude of 3 km and passes directly over a tracking telescope on the ground. when the angle of

elevation is π/6, this angle is decreasing at a rate of π/4 rad/min. how fast is the plane traveling at that time?
Physics
1 answer:
Lubov Fominskaja [6]3 years ago
3 0
The solution is:tan(θ) = opp / adj tan(θ) = y/x xtan(θ) = y 
Find x:
x = y/tan(θ) 
So x = 3/tan(π/6) 
Perform implicit differentiation to get the equation:
dx/dt * tan(θ) + x * sec²(θ) * dθ/dt = dy/dt 
Since altitude remains the same, dy/dt = 0. Now... 
dx/dt * tan(π/6) + 3/tan(π/6) * sec²(π/4) * -π/4 = 0 
changing the equation, will give us:
dx/dt = [3/tan(π/6) * sec²(π/6) * π/4} / tan(π/6) ≈ 12.83 km/min 
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Tresset [83]

Answer: 2.92 s

Explanation:

Given

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We know, change in momentum is Impulse

\Rightarrow F\cdot \Delta t=m(\Delta v)

\Rightarrow 25.9\cdot \Delta t=3.9(15.9-(-3.5))\\\\\Rightarrow \Delta t=\dfrac{3.9\times 19.4}{25.9}=2.92\ s

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Answer:

4.44 rpm

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The acceleration due to gravity is given by

g=\frac{GM}{r^2}\\\Rightarrow g=\frac{6.67\times 10^{-11}\times 4.8\times 10^{22}}{\left(\frac{3138000}{2}\right)^2}\\\Rightarrow g=1.3\ m/s^2

Here the centripetal acceleration of the arm and acceleration due to gravity are equal

a_c=\omega^2R

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1\ rad/s=\frac{60}{2\pi}\ rpm

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barxatty [35]

Answer:

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