Answer:
Explanation:
From the given information:
The speed of a synchronous motor in relation to its frequency can be represented with the formula:

where,
the electrical frequency
is measured in Hz
the number of poles = P
For us to estimate the number of poles to have 50 Hz - 60 Hz Power, then we need to relate the frequencies of the above equation.
i.e

Thus, we can conclude that 10 poles synchronous motor is attached with 12 poles synchronous generator in order to convert 50 Hz to 60 Hz power.
Answer:
A: 1.962
B: 3.924
Explanation:
g = G *M /R^2
g = 9.807*M/R^2 the gravitational constant of ground level on earth is about 9.807
g = 9.807*5lbs/R^2 the average brick is about 5 pounds.
g = 9.807*5*10^2. I'm assuming the height is around ten feet to help you out.
with these numbers plugged in you get an acceleration of 0.4905 a final velocity after 4 seconds 1.962. It's height fallen after 4 seconds is 3.924.
( M = whatever the brick weighs it's not specified in the question)
(R = the distance from the ground or how high the scaffold is)
(hopefully you can just plug your numbers in there hope this helps)
Answer:
Explanation:
E = σ/ε = (F/A) / (ΔL/L)
E = (mg/(πd²/4) / (ΔL/L)
E = (4mg/(πd²) / (ΔL/L)
E = 4Lmg/(πd²ΔL)
E = 4(30.0)(90)(9.8)/(π(0.01²)0.25)
E = 1.35 x 10⁹ Pa or 1.35 GPa
Answer:
(a) A = 1 mm
(b) 
(c) ![a_{max}=606.4 m/s^{2}/tex]Explanation:Distance moved back and forth = 2 mm Frequency, f = 124 HzSo, amplitude is the half of the distance traveled back and forth. (a) So, amplitude, A = 1 mm(b) Angular frequency, ω = 2 π f = 2 x 3.14 x 124 = 778.72 rad/s The formula for the maximum speed is given by [tex]V_{max}=\omega \times A](https://tex.z-dn.net/?f=a_%7Bmax%7D%3D606.4%20m%2Fs%5E%7B2%7D%2Ftex%5D%3C%2Fp%3E%3Cp%3E%3Cstrong%3EExplanation%3A%3C%2Fstrong%3E%3C%2Fp%3E%3Cp%3EDistance%20moved%20back%20and%20forth%20%3D%202%20mm%20%3C%2Fp%3E%3Cp%3EFrequency%2C%20f%20%3D%20124%20Hz%3C%2Fp%3E%3Cp%3ESo%2C%20amplitude%20is%20the%20half%20of%20the%20distance%20traveled%20back%20and%20forth.%20%3C%2Fp%3E%3Cp%3E%28a%29%20So%2C%3Cstrong%3E%20amplitude%2C%20A%20%3D%201%20mm%3C%2Fstrong%3E%3C%2Fp%3E%3Cp%3E%28b%29%20Angular%20frequency%2C%20%CF%89%20%3D%202%20%CF%80%20f%20%3D%202%20x%203.14%20x%20124%20%3D%20778.72%20rad%2Fs%20%3C%2Fp%3E%3Cp%3EThe%20formula%20for%20the%20maximum%20speed%20is%20given%20by%20%3C%2Fp%3E%3Cp%3E%5Btex%5DV_%7Bmax%7D%3D%5Comega%20%5Ctimes%20A)


(c) The formula for the maximum acceleration is given by


[tex]a_{max}=606.4 m/s^{2}/tex]