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Ksju [112]
3 years ago
8

Two different electrical devices have the same power consumption, but one is meant to be operated on 120-V AC and the other on 2

40-V AC. What is the ratio of the resistance of 240V device to the resistance of the 120V device i.e. \frac{R_{240}}{R_{120}} ​R ​120 ​​ ​ ​R ​240 ​​ ​​ ?
Physics
1 answer:
kupik [55]3 years ago
8 0

Answer:4

Explanation:

For same Power one device is operated at V_1=120 V and another at V_2=240\ V

Suppose R_1 and R_2 is the resistance of first and second device respectively

using P=\frac{V^}{R}

For first device

P=\frac{(120)^2}{R_1}

R_1=\frac{120^2}{P}

For second Device

R_2=\frac{240^}{P}

Ratio of resistances are

\frac{R_2}{R_1}=\frac{240^2}{120^2}

\frac{R_2}{R_1}=(2)^2

\frac{R_2}{R_1}=4

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A test charge of -3x10^-7 C is located 7 cm to the right of a charge of -9x10^-6 C and 20 cm to the left of a charge of +10x10^-
musickatia [10]

Answer:

5.634 N rightwards

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qo = - 3 x 10^-7 C

q1 = - 9 x 10^-6 C

q2 = 10 x 10^-6 C

r1 = 7 cm = 0.07 m

r2 = 20 cm = 0.2 m

The force on test charge due to q1 is F1 which is acting towards right

According to the Coulomb's law

F_{1}=\frac{Kq_{1}q_{0}}{r_{1}^{2}}

F1 = (9 x 10^9 x 9 x 10^-6 x 3 x 10^-7) / (0.07 x 0.07)

F1 = 4.959 N rightwards

The force on test charge due to q2 is F1 which is acting towards right

According to the Coulomb's law

F_{2}=\frac{Kq_{2}q_{0}}{r_{2}^{2}}

F2 = (9 x 10^9 x 10 x 10^-6 x 3 x 10^-7) / (0.2 x 0.2)

F2 = 0.675 N rightwards

Net force on the test charge

F = F1 + F2 = 4.959 + 0.675 = 5.634 N rightwards

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