Common transparent tape becomes charged when pulled from a dispenser. If one piece is placed above another, the repulsive force
1 answer:
Answer:
q = 2 10⁻⁸ C
Explanation:
For this exercise we use the translational equilibrium equation
F_e -A =
F_e = W
the electric force is given by Coulomb's law
F_e =
in this case they indicate that the loads on the tapes are equal
F_e = k q² / r²
we substitute
k q² / r² = m g
q =
calculate
q =
q =
q = 1,999 10⁻⁸ C
q = 2 10⁻⁸ C
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Answer:
A = 2.4 A
Explanation:
The voltage in the circuit, V = 12 V
The given circuit shows four resistors with R₁ and R₂ arranged in series with both in parallel to R₃ and R₄ which are is series to each other
R₁ = 4 Ω
R₂ = 6 Ω
R₄ = 5 Ω
The voltage across R₃ = 6 V
Voltage across parallel resistors are equal, therefore;
The total voltage across R₃ and R₄ = 12 V
The total voltage across R₁ and R₂ = 12 V
The voltage across R₃ + The voltage across R₄ = 12 V
∴ The voltage across R₄ = 12 V - 6 V = 6 V
The current flowing through R₄ = 6V/(5 Ω) = 1.2 A
The current flowing through R₃ = The current flowing through R₄ = 1.2 A
The resistor, R₃ = 6 V/1.2 A = 5 Ω
Therefore, we have;
The sum of resistors in series are R₁ + R₂ and R₃ + R₄, which gives;
= R₁ + R₂ = 4 Ω + 6 Ω = 10 Ω
= R₃ + R₄ = 5 Ω + 5 Ω = 10 Ω
The sum of the resistors in parallel is given as follows;
Therefore;
Therefore;
The value of the current, <em>A</em>, in the circuit, I = V/
A = I = 12 V/(5 Ω) = 2.4 A
A = 2.4 A