(89000/102000)×100
=87.25%
(92000/104000)×100
=88.46%
efficiency is (output/input)×100
if u get confused which way input and output should go, remember the smaller value is always output and it's above in the fraction, then only it's possible to get a efficiency lower than 100.
Answer: to only change one factor in an experiment or test
Some metals having unpaired electrons contain a strong magnetic response, i.e, they can be magnetized by an external magnetic field.
-identifies an electric charge
-it can identify its polarity (positive or negative) if you compare it to a charge that you already know
-can identify the magnitude of a charge (how big of a charge it is)
Answer:
<h2>A.
6pF</h2>
Explanation:
If unknown capacitance C1, C2, C3 and C4 are connected in series to one another, their equivalent capacitance of the circuit will be expressed as shown

Given the capacitance's 3.0 pF, 2.0 pF, 5.0 pF and X pF connected in series to each other. If the equivalent capacitance of the circuit is 0.83 pF, then to get X, we will apply the formula above;

C₄ ≈ 6pF
Hence the value of the X capacitor is approximately 6pF