Answer:
(B) The total internal energy of the helium is 4888.6 Joules
(C) The total work done by the helium is 2959.25 Joules
(D) The final volume of the helium is 0.066 cubic meter
Explanation:
(B) ∆U = P(V2 - V1)
From ideal gas equation, PV = nRT
T1 = 21°C = 294K, V1 = 0.033m^3, n = 2moles, V2 = 2× 0.033=0.066m^3
P = nRT ÷ V = (2×8.314×294) ÷ 0.033 = 148140.4 Pascal
∆U = 148140.4(0.066 - 0.033) = 4888.6 Joules
(C) P2 = P1(V1÷V2)^1.4 =148140.4(0.033÷0.066)^1.4= 148140.4×0.379=56134.7 Pascal
Assuming a closed system
(C) Wc = (P1V1 - P2V2) ÷ 0.4 = (148140.4×0.033 - 56134.7×0.066) ÷ 0.4 = (4888.6 - 3704.9) ÷ 0.4 = 1183.7 ÷ 0.4 = 2959.25 Joules
(C) Final volume = 2×initial volume = 2×0.033= 0.066 cubic meter
Refer to the diagram shown below.
i = the current in the circuit., A
R₁ = the internal resistance of the battery, Ω
R₂ = the resistance of the 60 W load, Ω
Because the resistance across the battery is 8.5 V instead of 9.0 V, therefore
(R₁ )(i A) = 9 - 8.5 = (0.5 V)
R₁*i = 0.5 (10
Also,
R₂*i = 9.5 (2)
Because the power dissipated by R₂ is 60 W, therefore
i²R₂ = 60
From (2), obtain
i*9.5 = 60
i = 6.3158 A
From (1), obtain
6.3158*R₁ = 0.5
R₁ = 0.5/6.3158 = 0.0792 Ω = 0.08 Ω (nearest hundredth)
Answer: 0.08 Ω
Answer:
163.8 ft
Explanation:
In triangle ABD
= 155 ft


Using Pythagorean theorem in triangle ADC

= distance between the anchor points
distance between the anchor points is given as

Answer:
8.97 Watt
Explanation:
Resistance, R = 20 ohm
Inductance, L = 10 mH
V(t) = 20 Cos (1000 t + 45°)
Compare with the standard equation
V(t) = Vo Cos(ωt + Ф)
Ф = 45°
ω = 1000 rad/s
Vo = 20 V
Inductive reactance, XL = ωL = 1000 x 0.01 = 10 ohm
impedance is Z.


Z = 22.36 ohm



Apparent power is given by
P = Vrms x Irms
P = 14.144 x 0.634
P = 8.97 Watt