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Lerok [7]
3 years ago
8

What voltage is delivered to a 120-volt/5,000-watt load that is fed with #10 awg wire (1.24 ohms/1,000 feet) and located 750 fee

t from the 120-volt source? (round the final answer to two decimal places.)?
Physics
1 answer:
Zepler [3.9K]3 years ago
4 0

Answer:

It is given that power = 5000 W for 120 V and resistance is 1.24\Omega per 1000 ft of wire. The wire is 750 ft away from 120 V source. We need find the voltage delivered to this load.

Power, P=Voltage(E)\times Current(I)

\Rightarrow I=\frac{P}{E}=\frac{5000W}{120V}=41.67 A

Resistance of 750 ft wire, =R=\frac{1.24\Omega}{1000 ft}\times750 ft=0.93\Omega

Using Ohm's Law:

Voltage delivered to the load, V=IR

\Rightarrow V=41.67A\times0.93\Omega=38.75 V


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Given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.

Given the data in the question;

Hubble's constant; H_0 = 51km/s/Mly

Age of the universe; t = \ ?

We know that, the reciprocal of the Hubble's constant ( H_0 ) gives an estimate of the age of the universe ( t ). It is expressed as:

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H_0 = 51\frac{km}{\frac{s}{Mly} } = 51000\frac{m}{s\ *\ Mly}  \\\\H_0 = 51000\frac{m}{s\ *\ (9.46*10^{21}m)} \\\\H_0 =  5.39 *10^{-18}s^{-1}\\

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Age\ of\ Universe; t = \frac{1}{H_0}\\\\t = \frac{1}{ 5.39 *10^{-18}s^{-1}} \\\\t = 1.855 * 10^{17}s\\\\We\ convert\ to\ years\\\\t =  \frac{ 1.855 * 10^{17}}{60*60*24*365}yrs \\\\t = \frac{ 1.855 * 10^{17}}{31536000}yrs\\\\t = 5.88 *10^9 years

Therefore, given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.

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