Answer:
Three ways that engineers explore possible solutions in their projects are;
1) Prototyping
2) Simulation
3) Calculations
Explanation:
1) Prototyping is the process of experimental testing of samples of design, or model of a product with the possibility of the inclusion of control of parameters in order to determine the workability of a solution.
2) Simulation is the process of creating an imitation of a situation, operation or process which can be used to determine if a particular solution will be able to work as required in the simulated environment of a problem.
3) Calculations are used to find preliminary results of particular situations, their cause and effects based on scientific laws, theories and hypothesis such that the factor of the problem is equated with the available ideas to find the best possible solution.
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:
Clouds
Explanation:
It is created by trapped dust and water.
...simplify devices, reducing weight and the chance of failure.
Answer:
Explanation:
Our values are,
State 1
We know moreover for the tables A-15 that
State 2
For tables we know at T=320K
We need to use the ideal gas equation to estimate the mass, so
Using now for the final mass:
We only need to apply a energy balance equation:
The negative value indidicates heat ransfer from the system
