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sertanlavr [38]
3 years ago
12

1. Household circuits are often wired with two different widths of wires: 12-gauge and 14-gauge. The 12-gauge wire has a diamete

r of 1/12 inch while the 14-gauge wire has a diameter of 1/14 inch. Thus, 12-gauge wire has a wider cross section than 14-gauge wire. A 20-Amp circuit used for wall receptacles should be wired using 12-gauge wire and a 15-Amp circuit used for lighting and fan circuits should be wired using 14-gauge wire. Explain the physics behind such an electrical code.
Physics
2 answers:
sukhopar [10]3 years ago
8 0

Answer:

It can be said that a 12 gauge wire is wider than a 14 gauge wire, this means that it will have less resistance. This means that it allows the electric charge to flow at a much higher speed, allowing for a higher current.

Explanation:

As a conclusion, the 12 gauge wire is used in circuits that find 20A fuse protection. However, the 14 gauge wire can withstand less electrical current because it has a higher resistance. These 14 gauge cables are used in circuits that meet 15A fuse protection.

Jet001 [13]3 years ago
5 0

Answer:

This formula R =ρL/A

Where R = resistance of wire, ρ = resistivity of the wire and A = area of the wire. Shows there is an inverse relationship between Resistance and Area of the wire.

Explanation:

A simple way to explain the physics behind such an electrical code is to compare the flow of current through wires to the flow of water through pipes, they are similar in any respect. The resistance to the flow of current in an electric circuit is similar to the frictional experienced by water when flowing through water pipes. Just as water will flow easily with little resistance through a water pipe with the larger cross-sectional area than one with a smaller cross-sectional area, in the same way, wires with larger cross-sectional area will allow the flow of larger amount of current compared to wires with smaller cross-sectional area assuming all other variables are the same.

From the formula R =ρL/A

Where R = resistance of wire, ρ = resistivity of the wire and A = area of the wire

We can see that the resistance and area of the wire have an inverse relationship. An increase in the area of the wire will lead to a decrease in the resistance of the wire.

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7 0
3 years ago
An electron in an atom has an uncertainty of 0.2 nm. If it is doubled to 0.4 nm by what factor does the uncertainty in momentum
fenix001 [56]

Answer:

The uncertainty in momentum changes by a factor of 1/2.

Explanation:

By Heisenberg's uncertainty principle, ΔpΔx ≥ h/2π where Δp = uncertainty in momentum and Δx = uncertainty in position = 0.2 nm. The uncertainty in momentum is thus Δp ≥ h/2πΔx. If the uncertainty in position is doubled, that is Δx₁ = 2Δx = 0.4 nm, the uncertainty in momentum Δp₁ now becomes Δp₁ ≥ h/2πΔx₁ = h/2π(2Δx) = (h/2πΔx)/2 = Δp/2.

So, the uncertainty in momentum changes by a factor of 1/2.

4 0
3 years ago
¿Cuál es la frecuencia de una nota musical cuyo periodo es 0,005 s?
AlekseyPX

Answer:

La respuesta sería 200Hz

3 0
3 years ago
A nonconducting sphere of diameter 10.0 cm carries charge distributed uniformly inside with charge density of +5.50 µC/m3 . A pr
VLD [36.1K]

Answer:

t = 2.58*10^-6 s

Explanation:

For a nonconducting sphere you have that the value of the electric field, depends of the region:

rR:\\\\E=k\frac{Q}{r^2}

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

R: radius of the sphere = 10.0/2 = 5.0cm=0.005m

In this case you can assume that the proton is in the region for r > R. Furthermore you use the secon Newton law in order to find the acceleration of the proton produced by the force:

F=m_pa\\\\qE=m_pa\\\\k\frac{qQ}{r^2}=m_pa\\\\a=k\frac{qQ}{m_pr^2}

Due to the proton is just outside the surface you can use r=R and calculate the acceleration. Also, you take into account the charge density of the sphere in order to compute the total charge:

Q=\rho V=(5.5*10^{-6}C/m^3)(\frac{4}{3}\pi(0.05m)^3)=2.87*10^{-9}C\\\\a=(8.98*10^9Nm^2/C^2)\frac{(1.6*10^{-19}C)(2.87*10^{-9}C)}{(1.67*10^{-27}kg)(0.05m)^2}=9.87*10^{11}\frac{m}{s^2}

with this values of a you can use the following formula:

a=\frac{v-v_o}{t}\\\\t=\frac{v-v_o}{a}=\frac{2550*10^3m/s-0m/s}{9.87*10^{11}m/s^2}=2.58*10^{-6}s

hence, the time that the proton takes to reach a speed of 2550km is 2.58*10^-6 s

3 0
3 years ago
A harmonic wave on a string with a mass per unit length of 0.050 kg/m and a tension of 60 N has an amplitude of 5.0 cm. Each sec
Dennis_Churaev [7]

Answer:

Power of the string wave will be equal to 5.464 watt

Explanation:

We have given mass per unit length is 0.050 kg/m

Tension in the string T = 60 N

Amplitude of the wave A = 5 cm = 0.05 m

Frequency f = 8 Hz

So angular frequency \omega =2\pi f=2\times 3.14\times 8=50.24rad/sec

Velocity of the string wave is equal to v=\sqrt{\frac{T}{\mu }}=\sqrt{\frac{60}{0.050}}=34.641m/sec

Power of wave propagation is equal to P=\frac{1}{2}\mu \omega ^2vA^2=\frac{1}{2}\times 0.050\times 50.24^2\times 34.641\times 0.05^2=5.464watt

So power of the wave will be equal to 5.464 watt

6 0
3 years ago
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