Given that the potential difference is V = 1.5 V.
The length of the wire is l = 1.5 m.
The cross-sectional area is
The resistivity of the wire is
We have to find the power dissipated in the wire.
First, we need to calculate resistance.
The resistance can be calculated as
The formula to calculate power is
Substituting the values, the power will be
Thus, the power dissipated in the wire is 17.1 W
Answer:
If its twice as far the light will be twice as bright.
Explanation:
The energy transferred by the appliance using mains electricity is 17.3 KJ
<h3>Data obtained from the question </h3>
- Potential difference (V) = 230V
- Charge (Q) = 150 C
<h3>How to determine the energy transferred </h3>
The energy transferred can be obtained as follow:
E = ½QV
E = ½ × 150 × 230
E = 75 × 230
E = 17250 J
Divide by 1000 to express in kilojoules
E = 17250 / 1000
E = 17.3 KJ
Learn more about energy stored in a capacitor:
brainly.com/question/14739936
<span>The Earth’s internal "((HEAT))" source provides the energy for our dynamic planet, providing it with the driving force for on-going disastrous events such as earthquakes and volcanic eruptions and for plate-tectonic motion. </span>