We now have an algebraic expression with only one variable, which can be solved. Once we have that, we can plug it back into one of the original equations (or the expression derived in Part A) to solve for the other variable. When this is done with the system of two equations from Parts A and B, what is the solution? Enter AA, then BB, as two numbers, separated by a comma.

Next a skier is pulled by a tow rope up a frictionless ski slope that makes an angle of 15 with the horizontal. the rope moves parallel to the slope with a constant speed of 0.69 m/s. the force of the rope does 800 3 of work on the skier as the skier moves a distance of 8.4 m up the incline. (a) if the rope moved with a constant speed of 2.2 m/s how much work would the force of the rope do on the skier as the skler moved a distance of 8.4 m up the incline? at what rate is the force of the rope doing work on the skier when the rope moves with a speed of (b) 0.69 m/s and (c) 2.2 m/s?

Explain the comparisons between nature and packingtown that appears on pages 81 and 82 answer

Apair of 10μf capacitors in a high-power laser are charged to 1.7 kv.a. what charge is stored in each capacitor? b. how much energy is stored in each capacitor?