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Aneli [31]
3 years ago
10

The battery has a voltage of 12V. R1 = R2 = 100Ω.

Physics
1 answer:
Margaret [11]3 years ago
4 0

Answer:

<h3> FOR PARALLEL CONNECTION</h3><h3>I1 = 0.12A</h3><h3>I2 = 0.12A </h3><h3>IT =0.24A</h3><h3>FOR SERIES CONNECTION</h3><h3>I1 = I2 = 0.06A</h3><h3>IT =0.06A</h3><h3 />

Explanation:

According to ohms law, V =ITRt

V is the supply voltage

IT is the total current flowing in the circuit

Rt is the total equivalent resistance

Given R1= R2= 100Ω

V= 12V

FOR PARALLEL CONNECTION;

To calculate the total current IT in the battery, we need to calculate the total equivalent resistance RT first. For a parallel connected circuit, the equivalent resistance in the circuit is the sum of the reciprocal of its individual resistances as shown;

\frac{1}{RT} = \frac{1}{100} + \frac{1}{100}\\\frac{1}{RT} = \frac{2}{100} \\\frac{RT}{1} = \frac{100}{2}\\

RT = 50Ω

from the equation above;

IT = V/RT

IT = 12/50

IT = 0.24A

Note that in a parallel connected circuit, different current flows through the resistances but the same voltage is across them.

IT = I1+I2

For current in resistance R1;

I1 = V/R1

I1 = 12/100

I1 = 0.12A

Since both resitance are the same, they will share the total current equally. Therefore I2 = 0.12A

FOR SERIES CONNECTION;

The total equivalent resistance in the circuit will be the sum of their individual resistances.

RT = R1+ R2

RT = 100Ω+100Ω

RT = 200Ω

IT = V/RT

IT = 12/200

IT = 0.06A

Since the resistances are connected in series, the same current will flow through them but different voltages. The total current flowing in the circuit will be the same current flowing through the resistors.

Therefore I1 = I2 = 0.06A

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3 years ago
If frequency is kept constant, how are the velocity and wavelength of a wave related?
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<h3><u>Answer;</u></h3>

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Answer:

a_M=1.92a_A

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r_A = Distance of Alex = 6 ft

Ratio of centripetal acceleration is given by

\dfrac{a_M}{a_A}=\dfrac{\omega_M^2r_M}{\omega_A^2r_A}\\\Rightarrow \dfrac{a_M}{a_A}=\dfrac{r_M}{r_A}\\\Rightarrow a_M=a_A\dfrac{r_M}{r_A}\\\Rightarrow a_M=\dfrac{11.5}{6}a_A\\\Rightarrow a_M=1.92a_A

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3 years ago
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Answer:

The transverse wave will travel with a speed of 25.5 m/s along the cable.

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let T = 2.96×10^4 N be the tension in in the steel cable, ρ  = 7860 kg/m^3 is the density of the steel and A = 4.49×10^-3 m^2 be the cross-sectional area of the cable.

then, if V is the volume of the cable:

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but V = A×L , where L is the length of the cable.

m = ρ×(A×L)

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then the speed of the wave in the cable is given by:

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Therefore, the transverse wave will travel with a speed of 25.5 m/s along the cable.

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