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.
Benthos
Option b is the answer
Answer:
Explanation:
Check attachment for solution
Explanation:
Period P has units of seconds (s).
Length has units of meters (m).
Mass has units of kilograms (kg).
Acceleration has units of meters per second squared (m/s²).
Dimensional analysis:
s = √(m / (m/s²))
Therefore:
P = k √(L/g)
where k is a dimensionless constant.
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