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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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galina1969 [7]

Given:

The force of attraction is F = 48.1 N

The separation between the charges is

\begin{gathered} r=\text{ 60.9 cm} \\ =\text{ 60.9}\times10^{-2}\text{ m} \end{gathered}

Also, the magnitude of charge q1 = q2 = q.

To find the magnitude of charge.

Explanation:

The magnitude of charge can be calculated by the formula

\begin{gathered} F=\frac{k(2q)}{r^2} \\ q=\frac{Fr^2}{2k} \end{gathered}

Here, k is the Coulomb's constant whose value is

k\text{ = 9}\times10^9\text{ N m}^2\text{ /C}^2

On substituting the values, the magnitude of charge will be

\begin{gathered} q=\frac{48.1\times(60.9^\times10^{-2})^2}{2\times9\times10^{^9}} \\ =9.91\times10^{-10}\text{ C} \\ =9.91\times10^{-4}\text{ }\mu C \end{gathered}

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1 year ago
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As per the question, the mass of meteorite [ m]= 50 kg

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When the meteorite falls on the ground, it will give whole of its kinetic energy to earth.

We are asked to calculate the gain in kinetic energy of earth.

The kinetic energy of meteorite is calculated as -

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                                                               =\frac{1}{2}50* 10^{6}\ J

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Here, J stands for Joule which is the S.I unit of energy.

Hence,\ the\ kinetic\ energy\ gained\ by\ earth\ is\ 25*10^6\ J

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Read 2 more answers
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