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:
3000 J
Explanation:
Kinetic energy is:
KE = ½ mv²
If m = 15 kg and v = -20 m/s:
KE = ½ (15 kg) (-20 m/s)²
KE = 3000 J
The electromagnetic force, and the gravitational force<span>.
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Responder:
<h2>
490 julios
</h2>
Explicación:
Se dice que el trabajo se realiza cuando una fuerza aplicada a un objeto hace que el objeto se mueva a través de una distancia. El trabajo realizado por un cuerpo se expresa mediante la fórmula;
Workdone = Fuerza * Distancia
Como Fuerza = masa * aceleración,
Workdone = masa * aceleración * distancia
Masa dada = 5.0kg, aceleración = 2.0m / s² d =?
Para obtener d, usaremos una de las leyes del movimiento,
d = ut + 1 / 2at²
u = 0 (ya que el cuerpo acelera desde el reposo) yt = 7.0s
d = 0 + 1/2 (2) (7) ²
d = 49m
Workdone = 5 * 2 * 49
Workdone = 490 Julios
Answer:
electrons
Explanation:
The magnitude of the electric field outside an electrically charged sphere is given by the equation
where
k is the Coulomb's constant
Q is the charge stored on the sphere
r is the distance (from the centre of the sphere) at which the field is calculated
In this problem, the cloud is assumed to be a charged sphere, so we have:
is the maximum electric field strength tolerated by the air before breakdown occurs
is the radius of the sphere
Re-arranging the equation for Q, we find the maximum charge that can be stored on the cloud:
Assuming that the cloud is negatively charged, then
And since the charge of one electron is
The number of excess electrons on the cloud is
