Carla draws two circuit diagrams that connect the same components in different ways, as shown.
2 answers:
Answer:
Circuit A is a series circuit
The total resistance in Circuit A is greater than that in Circuit B
Explanation:
The missing diagram is here attached. For circuit A, considering that in a series circuit, the current through each load is the same and the total voltage across the circuit is the sum of the voltages across each load. Moreover, current flowing through each resistor in a parallel circuit is different, depending on the resistance. Therefore, we can say that when resistors are connected in parallel, more current flows from the source than flow for any of them individually, so the total resistance is lower.
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The charge on each of the equally charged drops of hairspray willl be 7 × 10 ⁻¹³ C
<h3>What is Columb's law?</h3>The force of attraction between two charges, according to Coulomb's law, is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.
Similar charges repel each other, whereas charges that are opposed attract each other.
Given data;
Electric force,F = 9 × 10 ⁻⁹ N
Distance between charges,d = 7 × 10⁻⁴ m
Chrge,q₁ = q₂ =q C
From Columb's law;
Hence the charge on each of the equally charged drops of hairspray willl be 7 × 10 ⁻¹³ C
To learn more about Columb's law refer to the link;
#SPJ1
Answer:
4.04m
Explanation:
First you calculate the velocity of the block when it leaves the spring. You calculate this velocity by taking into account that the potential energy of the spring equals the kinetic energy of the block, that is:
To find the distance in which the block stops you use the following expression (net work done by the friction force is equal to the difference in the kinetic energy of the block):
where Ff is the friction force. By replacing the values of the parameters you obtain:
hence, the distance to the original position is 3.96m+0.08m=4.04m
Answer:
1) Ex(P) = -8.34602 N/C
2) = -5.23850174216 N/C
3) Question (3) is a similar question to (1)
4) = -5.23850174216 N/C
5) ≈ 1.041466 C/m²
6) σₐ ≈ 2.2330413 C/m²
Explanation:
The given parameter of the point charge located at the center of a conducting shell
The charge of the point charge, q₁ = -8.9 μC
The inner radius of the shell, a = 2.8 cm
The outer radius of the shell, b = 4.1 cm
The charge of the conducting shell, q₂ = 2.2 μC
Therefore, we have;
1) The point P(8.5, 0)
By plugging in the values, we have;
For
R₁ < R₂ < r, for the electric field at the point, 'P', we have;
Ex(P) = -8.34602 N/C
2) For the point given with coordinates (8.5, 0), the distance of the y-component of point from the center = 0
The y-component of the electric field = 0 N/C
4) For r = 1.4 cm, along the y-axis we have;
R₁ < r < R₂
Therefore, we have;
Substituting the values, we get;
= -5.23850174216 N/C
5) The charge density, , is given as follows;
6) Similarly, we have;