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.

A1,700kg car is being used to give a 1,400kg car a push start by exerting a force of 140n the impulse on the smaller car during the 30.0s of contact is +670kg*m/s. what is the impulse of the smaller car on the larger car? -814 kg*m/s 0kg *m/s -670kg*m/s -550kg*m/s

Find the equation for the plane through upper p 0 left parenthesis negative 4 comma negative 8 comma negative 5 right parenthesis perpendicular to the following line. xequalsnegative 4 minus t, yequalsnegative 8 plus 2 t, zequals3 t, minusinfinityless thantless thaninfinity

At a certain instant after jumping from the airplane a, a skydiver b is in the position shown and has reached a terminal (constant) speed vb = 52 m/s. the airplane has the same constant speed va = 52 m/s, and after a period of level flight is just beginning to follow the circular path shown of radius ρa = 2330 m. (a) determine the velocity and acceleration of the airplane relative to the skydiver. (b) determine the time rate of change of the speed vr of the airplane and the radius of curvature ρr of its path, both as observed by the nonrotating skydiver.