19. A mechanic used a hydraulic lift to raise a \( 12,054 \mathrm{~N} \) car 1.89 m above the floor of a garage. It took 4.75 s to raise the car. What was the power output of the lift? Type a response 20. How much work is done using a 500-watt microwave oven for 5 minutes? Type a response 20 Hen

**Problem 19** 1. The work done to lift the car is given by \[ W = F \times d \] where \( F = 12\,054 \, \mathrm{N} \) (force) and \( d = 1.89 \, \mathrm{m} \) (distance). Thus, \[ W = 12\,054 \times 1.89 \approx 22\,782.06 \, \mathrm{J} \] 2. The power output is the work done per unit time: \[ P = \frac{W}{t} \] Given \( t = 4.75 \, \mathrm{s} \), we have \[ P = \frac{22\,782.06}{4.75} \approx 4\,796.22 \, \mathrm{W} \] Rounding to an appropriate number of significant figures, the power output is about \[ P \approx 4.80 \times 10^3 \, \mathrm{W} \] --- **Problem 20** 1. The work done (or energy used) with a microwave oven is calculated using \[ W = P \times t \] where \( P = 500 \, \mathrm{W} \) and \( t = 5 \, \mathrm{minutes} \). First, convert the time into seconds: \[ t = 5 \times 60 = 300 \, \mathrm{s} \] 2. Now calculate the work: \[ W = 500 \times 300 = 150\,000 \, \mathrm{J} \] --- **Answers** - The power output of the lift is approximately \( 4.80 \times 10^3 \, \mathrm{W} \). - The work done by the 500-watt microwave oven in 5 minutes is \( 150\,000 \, \mathrm{J} \).