Flammable and combustible liquids themselves do not burn. It is the mixture of their vapours and air that burns. Gasoline, with a flashpoint of -40°C (-40°F), is a flammable liquid. Even at temperatures as low as -40°C (-40°F), it gives off enough vapour to form a burnable mixture in air.
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Lamp Wattage is utilized with a CU of 0.75 and 80 percent of the available light reaches the work surface, the remaining 20 percent is absorbed by walls and other objects in the space, resulting in an illumination level of 50 f-c over a 100 ft2 area
The term "lumen" refers to the emission of "luminous flux," which is a measurement of the total amount of visible light emitted by a source in a certain amount of time. The illumination level over a 100 ft2 area will be 50 f-c Lamp Wattage with a CU of 0.75 used, with 80 percent of the available light reaching the work surface and the remaining 20 percent being absorbed by walls and other objects in the room.
Total Lamp Wattage = Number of Lamps X Wattage of Each Lamp.
The fixtures' total wattage is 2 x 32, or 64 watts.
Lumen per Fixtures equals Lumen Efficiency (Lumen per Watt) times the Watt of each Fixture
85x 64 = 5440 lumens per fixture.
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Answer:
20 km/h
Explanation:
45 km ÷ 2.25 hours (15 mins is 0.25 hours)
= 20
20 km/h
Answer:
Explanation:
In order to convert the work function of cesium from electronvolts to Joules, we must use the following conversion factor:
In our problem, the work function of cesium is
so, we can convert it into Joules by using the following proportion:
