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dimulka [17.4K]
3 years ago
9

If a 6.1 A resistive load is connected by 2-conductor stranded 2-AWG copper wire to a source voltage of 125.2 V that is 358 ft a

way (one-way distance), then what is the voltage at the load
Physics
1 answer:
polet [3.4K]3 years ago
3 0

Answer:

124.86 V

Explanation:

We have to first calculate the voltage drop across the copper wire. The copper wire has a length of 358 ft

1 ft = 0.3048 m

358 ft = 109.12 m

The diameter of 2 AWG copper wire (d) = 6.544 mm = 0.006544 m

The area of the wire = πd²/4 = (π × 6.544²)/4 = 33.6 mm²

Resistivity of wire (ρ) = 0.0171 Ω.mm²/m

The resistance of the wire = \frac{\rho A}{l}=\frac{0.0171*109.12 }{33.6} =0.056\ ohm

The voltage drop across wire = current * resistance = 6.1 A * 0.056 ohm = 0.34 V

The voltage at end = 125.2 - 0.34 = 124.86 V

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330 grams of boiling water (temperature 100°C, specific heat capacity 4.2 J/K/gram) are poured into an aluminum pan whose mass i
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Answer:

T = 74°C

Explanation:

Given Mw = mass of water = 330g, Ma = mass of aluminium = 840g

Cw = 4.2gJ/g°C = specific heat capacity of water and Ca = 0.9J/g°C = specific heat capacity of aluminium

Initial temperature of water = 100°C.

Initial temperature of aluminium = 29°C

When the boiling water is poured into the aluminum pan, heat is exchanged and after a short time the water and aluminum pan both come to thermal equilibrium at a common temperature T.

Heat lost by water equal to the heat gained by aluminium pan.

Mw × Cw×(100 –T) = Ma × Ca × (T–29)

330×4.2×(100– T) = 890×0.9×(T–29)

1386(100 – T) = 801(T –29)

1386/801(100 – T) = T – 29

1.73(100 – T) = T – 29

173 –1.73T = T –29

173+29 = T + 1.73T

202 = 2.73T

T = 202/2.73

T = 74°C

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A U-tube is open to the atmosphere at both ends. Water is poured into the tube until the water column on the vertical sides of t
zmey [24]

Answer:

The value is  \Delta h  =  0.003 \  m

Explanation:

From the question we are told that

   The  height of the water is  h_1  =  10 \ cm  =  0.10 \  m

    The  density of  oil is \rho_o  =  950 \  kg/m^3

  The  height of  oil  is  h_2  =  6 \ cm  =  0.06 \  m

Given that both arms of the tube are open then the pressure on both side is the same

So  

      P_a =  P_b

=>   Here  

             P_a  =  P_z + \rho_w  *  g *  h

where  \rho_w is the density of water with value  \rho_w =  1000 \ kg/m^3

and  P_z is the atmospheric pressure

and  

        P_b  =  P_z + \rho_o  *  g *  h_2

=>   P_z + \rho_w  *  g * h =    P_z + \rho_o  *  g *  h_2

=>    \rho_w   * h  =    \rho_o  *  h_2

=>      h  =  \frac{950 * 0.06 }{1000}

=>      h  = 0.057 \ m

The  difference in height is evaluated as    

           \Delta h  =  0.06 - 0.057

          \Delta h  =  0.003 \  m

     

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A transformer has 1500 turns on the primary coil and 30 000 turns on the secondary coil. What is the potential difference across
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Explanation:

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