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The power that the light is able to utilize out of the supply is only 0.089 of the given.
Power utilized = (0.089)(22 W)
= 1.958 W
= 1.958 J/s
The energy required in this item is the product of the power utilized and the time. That is,
Energy = (1.958 J/s)(1 s) = 1.958 J
Thus, the light energy that the bulb is able to produce is approximately 1.958 J.
the answer is 0.284 lb/in3
Answer:
a) The x coordinate of the third mass is -1.562 meters.
b) The y coordinate of the third mass is -0.944 meters.
Explanation:
The center of mass of a system of particles (
), measured in meters, is defined by this weighted average:
(1)
Where:
- Mass of the i-th particle, measured in kilograms.
- Location of the i-th particle with respect to origin, measured in meters.
If we know that 
, 
, 
, 
, 
and 
, then the coordinates of the third particle are:
a) The x coordinate of the third mass is -1.562 meters.
b) The y coordinate of the third mass is -0.944 meters.
