Answer: 244.05 J
Explanation:
To find speed at 30 m above the ground use equation:
V²=Vo²-2Gs
V0=31.4m/s
s=30m
G=9.81m/s²
-----------------------
V²=31.4²-2*9.81*30
V²=985.96+588.6
V²=1574.56
V=39.68m/s ---speed of arrow on 30 m obove the ground
Use equation for kinetic enrgy:
Ke=mV²/2
m=0.155kg
V=39.68m/s
-------------------------
Ke=0.155kg*(39.68m/s)²/2
Ke=0.155*1574.5/2
Ke=244.05J
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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A Framework for K–12 Science Education: Practices, Crosscutting Concepts, and Core Ideas (Framework) recommends science education in grades K–12 be built around three major dimensions: science and engineering practices, crosscutting concepts that unify the study of science and engineering through their common application across fields, and core ideas in the major disciplines of natural science.
Meters ?? or just a variable
3 is the answer teeeeeeeeeheeeeeeeee
