An elevator weighing 2,200 lbs is raised one story in 20 seconds. if the distance between stories is 10 feet, the power developed by the elevator motor is ft • lb/s. the power of the motor required for the elevator is h. p.

To calculate for the power developed in the elevator motor in ft.lb/s, we multiply the distance and the weight of the elevator and divide the product by the time.
                       Power = (10 ft)(2200 lb) / 20 s = 1100 ft.lb/s

Next, convert the calculated value to HP. 
                              1100 ft.lb/s x (1 HP/ 550 ft.lb/s) = 2 HP

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