a mechanic uses a hydraulic lift to raise a 1,200 kg car 0.50 m off the ground . how much work does the lift do on the car

The work done is 5880 J

Explanation:

Recall the formula for work as force times distance, therefore in this case the force is that oppose to the gravitational force on the car(of magnitude m * g), and the distance is 0.5 meters.

Then:

Work = 1200 * 9.8 * 0.5 J = 5880 J

two charges are fixed in place on the x-axis as shown. the charge at x = 0 is negative and has magnitude q = 2 μcoulombs (1 μcoulomb = 10-6 coulombs). the charge at x = d, (d = +12 cm), is positive and has magnitude q = 4 μcoulombs. find a point x = xo along the x-axis at which the electric field is zero.

volts times amps 6 x 0.5 3w