Answer:
Explanation:
Hello!
To solve this problem you must follow the following steps, which are fully registered in the attached image.
1. Draw the complete outline of the problem.
2. Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)
through prior knowledge of two other properties.
3. Use temodynamic tables to find the density of water in state 1, by means of temperature and quality, with this value and volume we can find the mass.
3. Use thermodynamic tables to find the internal energy in state 1 and two using temperature and quality.
4. uses the first law of thermodynamics that states that the energy in a system is always conserved, replaces the previously found values and finds the work done.
5. draw the pV diagram using the 300F isothermal line
Answer:
Explanation:
f = 50.0 Hz, L = 0.650 H, π = 3.14
C = 4.80 μF, R = 301 Ω resistor. V = 120volts
XL = wL = 2πfL
= 2×3.14×50* 0.650
= 204.1 Ohm
Xc= 1/wC
Xc = 1/2πfC
Xc = 1/2×3.14×50×4.80μF
= 1/0.0015072
= 663.48Ohms
1. Total impedance, Z = sqrt (R^2 + (Xc-XL)^2)= √ 301^2+ (663.48Ohms - 204.1 Ohm)^2
√ 90601 + (459.38)^2
√ 90601+211029.98
√ 301630.9844
= 549.209
Z = 549.21Ohms
2. I=V/Z = 120/ 549.21Ohms =0.218Ampere
3. P=V×I = 120* 0.218 = 26.16Watt
Note that
I rms = Vrms/Xc
= 120/663.48Ohms
= 0.18086A
4. I(max) = I(rms) × √2
= 0.18086A × 1.4142
= 0.2557
= 0.256A
5. V=I(max) * XL
= 0.256A ×204.1
=52.2496
= 52.250volts
6. V=I(max) × Xc
= 0.256A × 663.48Ohms
= 169.85volts
7. Xc=XL
1/2πfC = 2πfL
1/2πfC = 2πf× 0.650
1/2×3.14×f×4.80μF = 2×3.14×f×0.650
1/6.28×f×4.8×10^-6 = 4.082f
1/0.000030144× f = 4.082×f
1 = 0.000030144×f×4.082×f
1 = 0.000123f^2
f^2 = 1/0.000123048
f^2 = 8126.922
f =√8126.922
f = 90.14 Hz
Answer:
Answer for the question is given in the attachment .
Explanation:
Answer:
I = 1205.69 Lx
Explanation:
The irradiation or intensity of the solar radiation on the earth is maximum for the vertical fire, with a value I₀
I = I₀ sin θ
in this case with the initial data we can calculate the initial irradiance
I₀ = 
I₀ = 1600 /sin 53
I₀ = 2003.42 lx
for when the angle is θ = 37º
I = 2003.42 sin 37
I = 1205.69 Lx
