Hi there! :)
Reference the diagram below for clarification.
1.
We must begin by knowing the following rules for resistors in series and parallel.
In series:
In parallel:
We can begin solving for the equivalent resistance of the two resistors in parallel using the parallel rules.
Now that we have reduced the parallel resistors to a 'single' resistor, we can add their equivalent resistance with the other resistor in parallel (15 Ohm) using series rules:
2.
We can use Ohm's law to solve for the current in the circuit.
3.
For resistors in series, both resistors receive the SAME current.
Therefore, the 15Ω resistor receives 6A, and the parallel COMBO (not each individual resistor, but the 5Ω equivalent when combined) receives 6A.
In this instance, since both of the resistors in parallel are equal, the current is SPLIT EQUALLY between the two. (Current in parallel ADDS UP). Therefore, an even split between 2 resistors of 6 A is <u>3A for each 10Ω resistor</u>.
4.
Since the 15.0 Ω resistor receives 6A, we can use Ohm's Law to solve for voltage.
Answer:
550 kg
Explanation:
mass = E / gh
= 33000/60
=550
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Explanation:
The electric field of an isolated charged parallel-plate capacitor is given by :
........(1)
Where
q is the electric charge
A is the area of cross section of parallel plate
It is clear from equation (1) that the electric field of a parallel plate capacitor is directly proportional to the charge on the plate and inversely proportional to the area of cross section of a plate.
So, the correct option is (E) i.e. "none of the above".
Answer:
q₃=5.3nC
Explanation:
First, we have to calculate the force exerted by the charges q₁ and q₂. To do this, we use the Coulomb's Law:
Since we know the net force, we can use this to calculate q₃. As q₁ is at the right side of q₃ and q₁ and q₃ have opposite signs, the force F₁₃ points to the right. In a similar way, as q₂ is at the left side of q₃, and q₂ and q₃ have equal signs, the force F₂₃ points to the right. That means that the resultant net force is the sum of these two forces:
In words, the value of q₃ must be 5.3nC.
90km per hour. Speed = distance÷time Speed= 225/2.5. Speed = 90Km per hour
