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mario62 [17]
3 years ago
11

If the resistance of an electric circuit is 12 ohms and the voltage in the circuit is 60 V, the current flowing through the circ

uit is...
A. 60 A.
B. 5 A.
C. 0.2 A.
D. 720 A.
Physics
2 answers:
OlgaM077 [116]3 years ago
8 0

Current = (voltage)/(resistance) = (60 v) / (12 ohms) = 5 amperes  (B)
AnnyKZ [126]3 years ago
7 0
The formula for calculating the current flowing in a circuit is:
I = V /R
Where:
I = current = ?
V = voltage = 60 V
R = resistance = 12 Ohms
I = V /R = 60/ 12 = 5
Therefore, the current flowing through the circuit is 5 A.
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How much work is required to compress 5.05 mol of air at 19.5°C and 1.00 atm to one-eleventh of the original volume by an isothe
Rus_ich [418]

Explanation:

(a)  For an isothermal process, work done is represented as follows.

             W = -nRT ln(\frac{V_{2}}{V_{1}})

Putting the given values into the above formula as follows.

        W = -nRT ln(\frac{V_{2}}{V_{1}})

             = - 5.05 mol \times 8.314 J/mol K \times (19.5 + 273) K \times ln (\frac{\frac{V_{1}}{11}}{V_{1}})

             = -12280.82 \times ln (0.09)

             = -12280.82 \times -2.41

             = 29596.78 J

or,         = 29.596 kJ       (as 1 kJ = 1000 J)

Therefore, the required work is 29.596 kJ.

(b) For an adiabatic process, work done is as follows.

         W = \frac{P_{1}V^{\gamma}_{1}(V^{1-\gamma}_{2} - V(1-\gamma)_{1})}{(1 - \gamma)}

              = \frac{-nRT_{1}(11^{\gamma - 1} - 1)}{1 - \gamma}

              = \frac{-5.05 \times 8.314 J/mol K \times 292.5 (11^{1.4 - 1} - 1)}{1 - 1.4}

              = 49.41 kJ

Therefore, work required to produce the same compression in an adiabatic process is 49.41 kJ.

(c)   We know that for an isothermal process,

               P_{1}V_{1} = P_{2}V_{2}

or,       P_{2} = \frac{P_{1}V_{1}}{V_{2}}

                    = 1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})

                    = 11 atm

Hence, the required pressure is 11 atm.

(d)   For adiabatic process,  

          P_{1}V^{\gamma}_{1} = P_{2}V^{\gamma}_{2}

or,       P_{2} = P_{1} (\frac{V_{1}}{V_{2}})^{1.4}

                    = 1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})^{1.4}

                    = 28.7 atm

Therefore, required pressure is 28.7 atm.

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What minimum heat is needed to bring 250 g of water at 20 ∘C to the boiling point and completely boil it away? The specific heat
Amiraneli [1.4K]

Answer:633.8 KJ

Explanation:

Given

mass of water\left ( m\right )=250gm

Initial temperature\left ( T_i\right )=20^{\circ}C

Final temperature \left ( T_f\right )=100^{\circ}C

Specific heat of water \left ( c \right )=4190 J/kg-k

heat of vaporization\left ( L\right )=22.6\times 10^5 J/kg

Heat required for process\left ( Q\right )=heat to raise water temperature from 20 to 100 +Heat to vapourize water completely

Q=mc\left ( T_f-T_i\right )+mL

Q=0.25\times 4190\times \left ( 100-20\right )+0.25\times 22\times 10^5

Q=\left ( 0.838+5.5\right )\times 10^5

Q=6.338\times 10^5J=633.8 KJ

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