Aplane flying horizontally at an altitude of 1 mi and a speed of 580 mi/h passes directly over a radar station. find the rate at which the distance from the plane to the station is increasing when it is 2 mi away from the station. round your answer to the nearest whole number.

 the tangent function of 2mi/5mi = 21.8 degrees
cosine of that angle = cos (21.8 degrees)
                                =0.928
 multiply 0.928 with 570 = 570*0.928
     Closing velocity         =  529.23 mph
As the plane gets closer to radar station, tangent function becomes infinity and the cosine function becomes zero
0*570 = 0 

Answer: a charge of -1.0  × 10⁻⁴ c must be placed  at x = 26 m.the sign of the charge will be the same sign as the  -4.0  µc charge.======electricstatic force between two point charges  is given by    where k is coulomb's constant,  8.99×10⁹  n·m²/c², q₁ and q₂ are the point charges, and r is the distance between the two point charges.assign positive to q₁let q₁ =  6.0 µc, q₂ = -4.0  µc, q₀ represent the charge at the origin, and q₃ represent this new charge to place.the charge at the origin q₀ would experience the following force f₀₁ from q₁, with r = 6.0 m (note that 1  µc = 10⁻⁶ c)    the force from q₂    the force from q₃    the net force on the origin charge, q₀, has to be zero for no electrostatic forcehaving an equation where k and q₀ can be divided out of the equation will allow us to solve for q₃    isolating:     a charge of -1.0  × 10⁻⁴ c must be placed  at x = 26 m.the sign of the charge will be the same as the  -4.0  µc charge

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