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joja [24]
4 years ago
11

Ask Your Teacher An electric utility company supplies a customer's house from the main power lines (120 V) with two copper wires

, each of which is 34.0 m long and has a resistance of 0.109 Ω per 300 m. (a) Find the potential difference at the customer's house for a load current of 116 A.
Physics
1 answer:
natta225 [31]4 years ago
4 0

Answer:

The potential difference at the customer's house is 117.1 V.

Explanation:

a) The potential difference at the customer's house can be calculated as follows:

\Delta V_{h} = \Delta V_{p} - \Delta V_{l}

<u>Where</u>:

V_{h}: is the potential difference at the customer's house

V_{p}: is the potential difference from the main power lines = 120 V

V_{l}: is the potential difference from the lines

\Delta V_{h} = \Delta V_{p} - IR

The resistance, R, is:

\frac{0.109 \Omega}{300 m}*2*34.0 m = 0.025 \Omega

Now, the potential difference at the customer's house is:

\Delta V_{h} = 120 V - 116A*0.025 \Omega = 117.1 V

Therefore, the potential difference at the customer's house is 117.1 V.

I hope it helps you!

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1. On the planet Microid, trains that run on air tracks much like the air track you used in lab are the principle means of trans
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a) the mass of the second car = 28000 kg

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Explanation:

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mass of the train car m_1 = 6300 kg

speed of the train car v_1 = 12.0 m/s

mass of the second moving car m_2 = ???

speed of the second moving car v_2 = 2.2 m/s

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they both move with a speed v_f = 4.00 m/s

a)

Using the conservation of momentum :

m_1v_1+m_2v_2 = (m_1 + m_2)v_f

(6300*12)+m_2(2.2) = (6300 + m_2)4

(75600)+m_2(2.2) = 25200 + 4m_2

75600 - 25200  = 4m_2 -2.2m_2

50400 = 1.8m_2

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b)

To determine the amount of kinetic energy that was lost in the collision;

we will need to find the difference between the kinetic energy before the collision and after the collision;

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K.E _{lost} = K.E_{i} - K.E _{f}

K.E_{i}  =  \frac{1}{2} m_1v_1^2 + \frac{1}{2}m_2v_2^2

K.E_{i}  =  \frac{1}{2} (6300)(12)^2 + \frac{1}{2}(28000)(2.2)^2

K.E_{i}  =  521360 \ J

K.E_{f}  = \frac{1}{2}(m_1 + m_2 ) v_f ^2

K.E_{f}  = \frac{1}{2}(6300 + 28000 ) (4)^2

K.E_{f}  =274400 \ J

Now;  the  kinetic energy that was lost in the collision is calculated as follows:

K.E _{lost} = K.E_{i} - K.E _{f}

K.E_{lost} = (521360-274400) \ J

K.E_{lost} =246960 \ J

K.E_{lost} =246.9 * 10 ^3 \ J

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3 years ago
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