what is most delicious fod in the philippines?
Answer:
Well there are a lot of delicious food in the philppines but my most favorite is the Lechon, Adobo, Sisig, Chicken Curry, Crispy pata and Sinigang
A car generator turns at 400 rpm (revolutions per minute) when the engine is idling. It has a rectangular coil with 300 turns of dimensions 5.00 cm by 6.11 cm that rotates in an adjustable magnetic field. What is the field strength needed to produce a 24.0 V peak emf?
Answer:
The field strength needed is 0.625 T
Explanation:
Given;
angular frequency, ω = 400 rpm = (2π /60) x (400) = 41.893 rad/s
area of the rectangular coil, A = L x B = 0.0611 x 0.05 = 0.003055 m²
number of tuns of the coil, N = 300 turns
peak emf = 24 V
The peak emf is given by;
emf₀ = NABω
B = (emf₀ ) / (NA ω)
B = (24) / (300 x 0.003055 x 41.893)
B = 0.625 T
Therefore, the field strength needed is 0.625 T
Estimate the rotor inertia assuming that the rotor is a cylinder of radius 8.98 mm, and length 25 mm, with a material of 100% copper. Explain why the rotor inertia may differ from these assumptions?
Answer:
The moment of inertia of the rotor is approximately [tex]1.105\times 10^{-6}[/tex] kilogram-square meters.
The rotor inertia may differ from these assumption due to differences in the shape of cross section.
Explanation:
We assume that rotor can be represented as a solid cylinder of radius [tex]r[/tex], length [tex]l[/tex], made of cooper ([tex]\rho = 8960\,\frac{kg}{m^{3}}[/tex]) and whose axis of rotation passes through its center of mass and is parallel to its cross section. By definition of Moment of Inertia and Theorem of Parallel Axes, the moment of inertia of the rotot is:
[tex]I = \frac{1}{4}\cdot \rho \cdot \left(\frac{\pi}{4} \right) \cdot R^{3}\cdot (3\cdot R^{2}+L^{2})[/tex]
[tex]I = \frac{\pi}{16}\cdot \rho \cdot R^{3}\cdot (3\cdot R^{2}+L^{2})[/tex] (Eq. 1)
Where:
[tex]\rho[/tex] - Density of copper, measured in kilograms per cubic meter.
[tex]R[/tex] - Radius of the rotor, measured in meters.
[tex]L[/tex] - Length of the rotor, measured in meters.
[tex]I[/tex] - Moment of inertia, measured in kilogram-square meters.
If we know that [tex]\rho = 8960\,\frac{kg}{m^{3}}[/tex], [tex]L = 25\times 10^{-3}\,m[/tex] and [tex]R = 8.98\times 10^{-3}\,m[/tex], the estimated moment of inertia of the rotor is:
[tex]I = \frac{\pi}{16}\cdot \left(8960\,\frac{kg}{m^{3}} \right)\cdot (8.98\times 10^{-3}\,m)^{3}\cdot [3\cdot (8.98\times 10^{-3}\,m)^{2}+(25\times 10^{-3}\,m)^{2}][/tex]
[tex]I \approx 1.105\times 10^{-6}\,kg\cdot m^{2}[/tex]
The moment of inertia of the rotor is approximately [tex]1.105\times 10^{-6}[/tex] kilogram-square meters.
From D'Alembert's Formula we know that net force of rigid bodies experimenting rotation equals the product of moment of inertia and angular acceleration. In this case, the purpose is minimizing moment of inertia and it is done by modifying the shape of the cross section so that rotor could be aerodynamically more efficient.
Before you calculate the moment capacity for a steel beam, you have to determine the classification of beam.
a. True
b. False
Answer:
True
Explanation:
True - because different classification of steel beam have different yield strength.
The moment capacity for a steel beam is given by;
M = Mn / Ωₙ
where;
M - the maximum moment acting on the beam
Ωₙ - is the Safety Factor for Elements in Bending = 1.67
Mn - nominal moment of the steel, given as
[tex]M_n = Z_x *f_y[/tex]
where;
Zₓ - the Plastic Section Modulus in the x or strong axis.
[tex]f_y -[/tex] is the Yield Strength of the Steel (A36W, A46 W or A50W)
A36W = 36 ksi
A46 W = 46 ksi
A50W = 50 ksi
Thus, before you calculate the moment capacity for a steel beam, you have to determine the classification of beam, for the yield strength of the steel beam.
