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Lerok [7]
2 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]2 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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Electric motors require electricity to move the motor parts and do work (like an electric fan).
Elecgric generators actually burn diesel fuel to spin a motor around and around and around to GENERATE, or MAKE, electricity that you can then use to power your fans and lights in your house.

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2 years ago
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A circular saw blade with radius 0.175 m starts from rest and turns in a vertical plane with a constant angular acceleration of
ANEK [815]

Answer:

The distance the piece travel in horizontally axis is

L=3.55m

Explanation:

a=2 \frac{rev}{s^{2}} \\h=0.820m\\r = 0.125 m
\\d=150rev

d= 155 rev = 155(2\pi ) = 310\pi rad

a= 2.0 \frac{rev}{s^{2} } = 2.0(2\pi )  = 4.0\pi \frac{rev}{s^{2} }

d=d_{i}+vo*t+\frac{1}{2}*a*t^{2} \\ di=0\\vo=0\\d=\frac{1}{2}*a*t^{2}\\t=\sqrt{\frac{2*d}{a}}\\t=\sqrt{\frac{2*310 rad}{4\frac{rad}{s^{2}}}} \\t=12.449

w=a*t\\w=4\frac{rad}{s^{2}}*12.449s\\ w=49.79 \frac{rad}{s}

Now the angular velocity is the blade speed so:

V=w*r\\V=49.79 \frac{rad}{s}*0.175m\\V=8.7 \frac{m}{s}

assuming no air friction effects affect blade piece:

time for blade piece to fall to floor

t=\sqrt{\frac{2*h}{g}}\\t=\sqrt{\frac{2*0.820m}{9.8\frac{m}{s^{2} } }}\\t=0.409s

Now is the same time the piece travel horizontally

L=t*V\\L=0.409s*8.7\frac{m}{s}\\L=3.55m

blade piece travels  HORIZONTALLY = (24.5)(0.397) = 9.73 m  ANS

6 0
2 years ago
The gravitational field strength at a distance R from the center of moon is gR. The satellite is moved to a new circular orbit t
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Answer:

g'=\frac{g__R}{4}

Explanation:

Given:

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  • Distance of the satellite from the center of the moon, h=2R

<u>Now as we know that the value of gravity of any heavenly body is at height h is given as:</u>

g'=g__{R}} \times \frac{R^2}{(2R)^2}

g'=\frac{g__R}{4}

∴The gravitational field strength will become one-fourth of what it is at the surface of the moon.

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2 years ago
The temperature and pressure at the surface of Mars during a Martian spring day were determined to be -50 °C and 900 Pa, respect
Sidana [21]

Answer:

T = 273 + (-50) = 273 – 50 = 223 K

R = 188.82 J / kg K for CO2

Density (Martian Atmosphere) = P / RT = 900 / 188.92 x 223 = 900 / 42129.16 = 0.0213 kg / m^{3}

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4 0
2 years ago
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1. A diffraction grating with 5.000 x 103 lines/cm is used to examine the sodium
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Answer:

0.0002°, 0.1691°, 0.338°

Explanation:

Difference between the two line = 5.97 * 10-⁸m

d = 1 / N

N = 5.0 * 10³

d = 2.0 * 10⁴m

nL = Nsin¤

For first order

588.995 * 10-⁹ = 2.0 * 10-⁴ sin ¤

Sin¤ = 2.944*10-³

¤ = sin-¹ 0.002944

¤ = 0.1687°

First order ¤ =

Sin-¹(589.592*-⁹ / 2.0 * 10-⁴)

Sin-¹ (0.002947) = 0.1689°

Angular separation = 0.1689 - 0.1687 = 0.0002°

Second order ¤ = sin-¹ [2 (589.59*10-⁹ / 2.0*10-⁴)] = sin-¹ (0.005895)

Second order ¤ = 0.3378°

Angular difference = 0.3378° - 0.1687° = 0.1691°

Third order ¤ = sin-¹ [3(589.59*10-⁹ /2.0*10-⁴] = 0.5067°

Angular difference = 0.5067° - 0.1687° = 0.338°

7 1
2 years ago
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