A wire is 0.92 m long and 1.2 mm2 in cross-sectional area. It carries a current of 5.0 A when a 2.2 V potential difference is ap
plied between its ends. Calculate the conductivity σ of the material of which this wire is made.
1 answer:
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Answer:
0.5 kW
Explanation:
The given parameters are;
Volume of tank = 1 m³
Pressure of air entering tank = 1 bar
Temperature of air = 27°C = 300.15 K
Temperature after heating = 477 °C = 750.15 K
V₂ = 1 m³
P₁V₁/T₁ = P₂V₂/T₂
P₁ = P₂
V₁ = T₁×V₂/T₂ = 300.15 * 1 /750.15 = 0.4 m³
For ideal gas, = 5/2×R = 5/2*0.287 = 0.7175 kJ
PV = NKT
N = PV/(KT) = 100000×1/(750.15×1.38×10⁻²³)
N = 9.66×10²⁴
Number of moles of air = 9.66×10²⁴/(6.02×10²³) = 16.05 moles
The average mass of one mole of air = 28.8 g
Therefore, the total mass = 28.8*16.05 = 462.135 g = 0.46 kg
∴ dQ = 0.46*0.7175*(750.15 - 300.15) = 149.211 kJ
The power input required = The rate of heat transfer = 149.211/(60*5)
The power input required = 0.49737 kW ≈ 0.5 kW.
Answer:
vB = - 0.176 m/s (↓-)
Explanation:
Given
(AB) = 0.75 m
(AB)' = 0.2 m/s
vA = 0.6 m/s
θ = 35°
vB = ?
We use the formulas
Sin θ = Sin 35° = (OA)/(AB) ⇒ (OA) = Sin 35°*(AB)
⇒ (OA) = Sin 35°*(0.75 m) = 0.43 m
Cos θ = Cos 35° = (OB)/(AB) ⇒ (OB) = Cos 35°*(AB)
⇒ (OB) = Cos 35°*(0.75 m) = 0.614 m
We apply Pythagoras' theorem as follows
(AB)² = (OA)² + (OB)²
We derive the equation
2*(AB)*(AB)' = 2*(OA)*vA + 2*(OB)*vB
⇒ (AB)*(AB)' = (OA)*vA + (OB)*vB
⇒ vB = ((AB)*(AB)' - (OA)*vA) / (OB)
then we have
⇒ vB = ((0.75 m)*(0.2 m/s) - (0.43 m)*(0.6 m/s) / (0.614 m)
⇒ vB = - 0.176 m/s (↓-)
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