The value of 4.9 kW is equal to 4900 watts. Use the following equation to find the amperage for your question. I = W/E, Amps = Watts / Volts.
49 centimeters is equal to 0.49 meters. You can convert centimeters to meters by dividing by 100 since there are 100 centimeters in 1 meter.
A 20 HP three-phase motor operating at 230 volts will draw approximately 45 amps. You would typically use a wire size of at least AWG 8 for this application. However, it's recommended to consult the National Electrical Code (NEC) and a qualified electrician to ensure compliance with local regulations and safe installation practices.
To convert liters to milliliters, you would multiply the number of liters by 1000. Therefore, 49 liters is equal to 49,000 milliliters.
4.9 cm is equal to 49 mm. To convert from centimeters to millimeters, you simply multiply the number of centimeters by 10.
You could run 30 low-energy lights, enough to light about three houses, or you could run about 10 laptop computers, or 2-3 tower-type computers. Also phone chargers, home audio and video recorders. Not high-power things like irons, convector heaters, kettles or cookers.
To convert 0.49 to percent multiply by 100: 0.49 × 100 = 49 %
Answer: 49 km = 30.4471 mi.
Answer: 49 ºF = 9.4 ºC
Divide by seven
49 inches
- 94/5 + 915/16 + 94/15 = -49/5 + 159/16 + 139/15 = -2352/240 + 2385/240 + 2224/240 = 2257/240 = 997/240
49ºF = 9.4ºC
49/50
You may need to step up the voltage to minimize the power loss / voltage dip at the end of the 800 feet depending on: 1. the type of cable used (the higher resistance, the large cables would be required to not do this) 2. the amperage that you will need to draw. The voltage drop in the wires is due to the I^2*R losses in the cable. If your cable has a resistance of 1 ohm over 800 feet, and you draw 5 amps (so your load resistance is R = V/R = 240 / 5 = 48 ohms), the actual current delivered will be (I = V / R = 240 / 49) 4.9 amps, and the voltage at the end of the 800 feet will be (240 - 4.9*1) 235 volts. voltage is probably fine for whatever equipment you're using. If used continuously, the 800 ft of cable would waste 210kWh of energy/yr (15-35 dollars/year). If you need 20 amps, the load resistance would be (240 / 20) 12 ohms, current delivered to the load would be (240/13) = 18.5, and the voltage at the end of the 800 feet would be (240 - 18.5*1) 222 volts. This might be too low, depending on the equipment you're using and if you're area has a naturally low voltage to begin with. If used continuously, the 800 ft of cable would waste ~3,000kWh of energy/yr (180-450 dollars/year). As your question is written, a straight yes or no cannot be given. I hope the above explains enough to give you the knowledge you need to determine this on your own.
Without more information you cannot.
5g
49 centimeters is equal to 0.49 meters. You can convert centimeters to meters by dividing by 100 since there are 100 centimeters in 1 meter.