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.
Answer:
D
Explanation:
As we progress and learn more, scientists make new discoveries that can contradict earlier discoveries, and since are technology is better today we are able to discover a lot more to add to or change our previous theories.
Answer:
0.4
Explanation:
Given that a particular inductor is connected to a circuit where it experiences a change in current of 0.8 amps every 0.10 sec. If the inductor has a self-inductance of 2.0 V, what is the inductance
Using the power formula
P = IV
Substitute all the parameters
P = 0.8 × 2
P = 1.6 W
But P = I^2 R
Substitute power and current
1.6 = 0.8^2 R
R = 1.6 / 0.64
R = 2.5 ohms
Inductance = reciprocal of resistance
Inductance = 1 / 2.5
Inductance = 0.4
Proton positive; electron negative; neutron no charge<span>. </span>The charge<span> on the proton and </span>electron<span> are exactly the same size but opposite. The same number of protons and </span>electrons<span> exactly cancel one another in a neutral atom.
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hoped it helped
