To calculate the power required for a radiographic exposure at 76 kV and 500 mA, you can use the formula: Power (P) = Voltage (V) × Current (I). Here, P = 76 kV × 500 mA = 76,000 volts × 0.5 amps = 38,000 watts, or 38 kW. Therefore, the power required for the radiographic exposure is 38 kW.
it costs about €500
A zener diode with a rating of 500 mW will pass 50 mA at 10 V. (Power = voltage times current)Note: The question appears mis stated, in that it states a rating of 500 MW, not 500 mW. To my knowledge, there is no zener with a rating of 500 MW.
A welding machine rated at 500 amperes uses around 120000 watts of power. This is used in very heavy industrial applications due to its large power draw.
A: NO it cannot 500va is 500 x 1 amp = 500 watts you need a 700va to work properly,
around $500 a pole and $2 a foot about $1000 for the pole mounted transformer poles need to be less than 100' apart
Depends on the size of the amplifier you are installing and the power required to run it. You can easily install a 500 watt amp with no problems.
166W
The minimum system memory required for the ATI Radeon HD 4890 is 500 Watt or greater power supply with two 75W 6-pin PCI Expresså¨ power connectors recommended.
Exposure to ammonia at 500 ppm can cause irritation of the eyes, nose, and throat, leading to symptoms such as coughing, shortness of breath, and watery eyes. Prolonged exposure can result in more serious respiratory issues and damage to the respiratory system. It is important to limit exposure to ammonia and ensure proper ventilation in areas where it is present.
500
the reading room required illuminance around 500.
audionic mini size laptop speakers ( USB based) 300 watts best output result. no external power required.
500 W power
The population of Silicon Power is 500.
250,000
Since the second world war the U.S.A was a super power, now in 500 years it is time another country be in power.
To find the time taken to perform 500 joules of work at a power of 25 watts, you divide the work by the power. In this case, 500 Joules / 25 watts = 20 seconds. Therefore, it would take 20 seconds to perform 500 Joules of work with 25 watts of power.