Answer:
In which direction does the current in circuit A flow?
counterclockwise
<h2>What is the power dissipated by the resistor of resistance R2 for circuit A, given that E=10 V, R1=300ohms, and R2=5000ohms?
</h2><h2>Calculate the power to two significant figures.</h2><h2>0.064W</h2><h2 /><h2>For what ratio of R1 and R2 would power dissipated by the resistor of resistance R2 be the same for circuit A and circuit B?</h2><h2>R1/R2 =
1
</h2><h2 /><h2>Under which of the following conditions would power dissipated by the resistance R2 in circuit A be bigger than that of circuit B?
</h2><h2>Some answer choices overlap; choose the most restrictive answer.</h2><h2>R2>R1</h2><h2>
</h2>
Explanation:
Answer:
No.
Explanation:
- According to Faraday's law, the induced emf in the circuit is given by :
, it is proportional to the rate of change of magnetic flux.
- In this case, a short piece of wire that is not attached to anything and move it up and down in a magnetic field. It means that the circuit is not completed here. It is an open circuit. For the induction of current, a circuit must be completed.
- Hence, no current will induce.
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
Explanation:
20 joule is your answer
Answer:
here
mass m =100kg
distance d=50m
acceleration due to gravity a =10m/s²
work =force×displacement
= ma/d=100×10/50=20joule