Akite 100 ft above the ground moves horizontally at a speed of 12 ft/s. at what rate is the angle (in radians) between the string and the horizontal decreasing when 200 ft of string have been let out?

So this involves right triangles. The height is always 100. Let the horizontal be x and the length of string be z. So we have x2 + 1002 = z2. Now take its derivative in terms of time to get 2x(dx/dt) = 2z(dz/dt) So at your specific moment z = 200, x = 100âš3 and dx/dt = +8 substituting, that makes dz/dt = 800âš3 / 200 or 4âš3. Part 2 sin a = 100/z = 100 z-1 . Now take the derivative in terms of t to get cos a (da./dt) = -100/ z2 (dz/dt) So we know z = 200, which makes this a 30-60-90 triangle, therefore a=30 degrees or Ď€/6 radians. Substitute to get cos (Ď€/6)(da/dt) = (-100/ 40000)(4âš3) âš3 / 2 (da/dt) = -âš3 / 100 da/dt = -1/50 radians

Answerthe force between the charges will be  f/9 (one-ninth of force f)explanationcoulomb's law relates the magnitudes  two point charges q₁ and q₂, the distance, r, between the charges, and the electrostatic force, f, between the two charges:     k represents coulomb's constant, k = 8.99×10⁹ n·m²/c².note that if we have r = 3r, then   

Conductionhope this .