This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
ScanSolve AI Answer
Step 1: Identify the given values and constants. • Volume of water, V = 5 m^3 • Time, t = 1 minute = 60 seconds • Height, h = 30 m • Efficiency, = 80\% = 0.80 • Density of water, = 1000 kg/m^3 (standard value) • Acceleration due to gravity, g = 9.81 m/s^2 (standard value) Step 2: Calculate the mass of water pumped per minute. The mass (m) of water can be found using its density () and volume (V): m = × V m = 1000 kg/m^3 × 5 m^3 m = 5000 kg Step 3: Calculate the useful power (power output) required to lift the water. Useful power (P_out) is the rate at which potential energy is gained by the water. P_out = (mgh)/(t) P_out = (5000 kg) × (9.81 m/s^2) × (30 m)60 s P_out = 1471500 J60 s P_out = 24525 W Step 4: Calculate the actual power required (power input) by the pump, considering its efficiency. Efficiency () is defined as the ratio of useful power output to the total power input (P_in): = P_outP_in Rearranging the formula to find P_in: P_in = P_out P_in = 24525 W0.80 P_in = 30656.25 W The power required to pump the water is 30656.25 W. That's 2 down. 3 left today — send the next one.