Answer:
0.304 L of Freon is needed
Explanation:
Q = mCT
Q is quantity of energy that must be removed = 47 BTU = 47×1055.06 = 49587.82 J
C is specific heat of Freon = 74 J/mol.K = 74 J/mol.K × 1 mol/120 g = 0.617 J/g.K
T is temperature in the area of Mars = 189 K
m = Q/CT = 49587.82/(0.617×189) = 452.23 g = 452.24/1000 = 0.45223 kg
Density of Freon = specific gravity of Freon × density of water = 1.49 × 1000 kg/m^3 = 1490 kg/m^3
Volume of Freon = mass/density = 0.45223/1490 = 0.000304 m^3 = 0.000304×1000 = 0.304 L
It would be 72cm bc u need to add up all the line in the back to
Answer:
a) the inductance of the coil is 6 mH
b) the emf generated in the coil is 18 mV
Explanation:
Given the data in the question;
N = 570 turns
diameter of tube d = 8.10 cm = 0.081 m
length of the wire-wrapped portion l = 35.0 cm = 0.35 m
a) the inductance of the coil (in mH)
inductance of solenoid
L = N²μA / l
A = πd²/4
so
L = N²μ(πd²/4) / l
L = N²μ(πd²) / 4l
we know that μ = 4π × 10⁻⁷ TmA⁻¹
we substitute
L = [(570)² × 4π × 10⁻⁷× ( π × (0.081)² )] / 4(0.35)
L = 0.00841549 / 1.4
L = 6 × 10⁻³ H
L = 6 × 10⁻³ × 1000 mH
L = 6 mH
Therefore, the inductance of the coil is 6 mH
b)
Emf ( ∈ ) = L di/dt
given that; di/dt = 3.00 A/sec
{∴ di = 3 - 0 = 3 and dt = 1 sec}
Emf ( ∈ ) = L di/dt
we substitute
⇒ 6 × 10⁻³ ( 3/1 )
= 18 × 10⁻³ V
= 18 × 10⁻³ × 1000
= 18 mV
Therefore, the emf generated in the coil is 18 mV
Answer:
18 kJ
Explanation:
Given:
Initial volume of air = 0.05 m³
Initial pressure = 60 kPa
Final volume = 0.2 m³
Final pressure = 180 kPa
Now,
the Work done by air will be calculated as:
Work Done = Average pressure × Change in volume
thus,
Average pressure = 
= 120 kPa
and,
Change in volume = Final volume - Initial Volume = 0.2 - 0.05 = 0.15 m³
Therefore,
the work done = 120 × 0.15 = 18 kJ
