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BartSMP [9]
3 years ago
7

A 120 resistor a 60 ohm resistor and a 40 ohm resistor are connected in parallel to a 120 volt power source. what is the current

running through the entire circuit?
Physics
1 answer:
larisa [96]3 years ago
6 0

Answer:

6 A

Explanation:

First of all, we need to calculate the equivalent resistance of the circuit. The three resistors are connected in parallel, so their equivalent resistance is given by:

\frac{1}{R_T}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}=\frac{1}{120 \Omega}+\frac{1}{60 \Omega}+\frac{1}{40 \Omega}=\frac{3+2+1}{120 \Omega}=\frac{6}{120 \Omega}\\R_T = \frac{120}{6} \Omega

And now we can use Ohm's law to find the current in the circuit:

V=R_T II=\frac{V}{R_T}=\frac{120 V}{\frac{120}{6}\Omega}=6 A

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Dima020 [189]

Answer:

False

Explanation:

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||A||=\sqrt{A_x^2+A_y^2}

The magnitude of anything is never negative. It can be even seen from the formula that the components are squared. A squared value can never be negative. Even if the component is negative the square will be always positive.

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Parallel light rays with a wavelength of 563 nm fall on a single slit. On a screen 3.30 m away, the distance between the first d
MrMuchimi

Answer:

The width of the slit is 0.4 mm (0.00040 m).

Explanation:

From the Young's interference expression, we have;

(λ ÷ d) = (Δy ÷ D)

where λ is the wavelength of the light, D is the distance of the slit to the screen, d is the width of slit and Δy is the fringe separation.

Thus,

d = (Dλ) ÷ Δy

D = 3.30 m, Δy = 4.7 mm (0.0047 m) and λ = 563 nm (563 ×10^{-9} m)

d = (3.30 × 563 ×10^{-9} ) ÷ (0.0047)

  = 1.8579 × 10^{-6} ÷ 0.0047

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d = 0.00040 m

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What is the electric potential energy of an electron at the negative end of the cable, relative to the positive end of the cable
VashaNatasha [74]

Answer:

Electric potential energy at the negative terminal: 1.92\cdot 10^{-18}J

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In this problem, we know that:

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q=-1.60\cdot 10^{-19}C

We also know that the positive terminal is at potential

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Therefore, the potential difference (final minus initial) is

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This means that the electron when it is at the negative terminal has 1.92\cdot 10^{-18}J of energy more than when it is at the positive terminal.

Since the potential at the positive terminal is 0, this means that the electric potential energy of the electron at the negative end is

1.92\cdot 10^{-18}J

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