Answer: Voltage, V=120V; Power,P=10kW or 10,000W
P = V*I (where I is current)
∴ I = P/V =10000/120 = 83.3A
Resistance, R = V/I = 1.44 ohms
Power Factor(cosФ)= 0.9 = Resistance, R/Impedance, Z
∴ Z = 1.44/0.9 = 1.6 ohms
Explanation: The three-phase inductive load supply carries alternating current of the same frequency and voltage amplitude relative to a common reference but phase difference of one third of a cycle between each of three voltages and currents.
Answer:
A south and north pole (negative and positive pole)
Explanation:
Because opposites attract
Answer:
a) 43.20V
b) 2.71W/s
c) 40.25s
d) 7.77Nm
Explanation:
(a) The emf of a rotating coil with N turns is given by:
![emf=NBA\omega sin(\omega t)](https://tex.z-dn.net/?f=emf%3DNBA%5Comega%20sin%28%5Comega%20t%29)
N: turns
B: magnitude of the magnetic field
A: area
w: angular velocity
the emf max is given by:
![emf_{max}=NBA\omega=(50)(1.80T)(0.200m*0.100m)(24.0rad/s)\\\\emf_{max}=43.20V](https://tex.z-dn.net/?f=emf_%7Bmax%7D%3DNBA%5Comega%3D%2850%29%281.80T%29%280.200m%2A0.100m%29%2824.0rad%2Fs%29%5C%5C%5C%5Cemf_%7Bmax%7D%3D43.20V)
(b) the maximum rate of change of the magnetic flux is given by:
![\frac{d\Phi_B}{dt}=\frac{d(A\cdot B)}{dt}=\frac{d}{dt}(ABcos\omega t)=AB\omega sin(\omega t)\\\\\frac{d\Phi_B}{dt}_{max}=(\pi(0.200*0.100))(1.80T)(24.0rad/s)=2.71\frac{W}{s}](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5CPhi_B%7D%7Bdt%7D%3D%5Cfrac%7Bd%28A%5Ccdot%20B%29%7D%7Bdt%7D%3D%5Cfrac%7Bd%7D%7Bdt%7D%28ABcos%5Comega%20t%29%3DAB%5Comega%20sin%28%5Comega%20t%29%5C%5C%5C%5C%5Cfrac%7Bd%5CPhi_B%7D%7Bdt%7D_%7Bmax%7D%3D%28%5Cpi%280.200%2A0.100%29%29%281.80T%29%2824.0rad%2Fs%29%3D2.71%5Cfrac%7BW%7D%7Bs%7D)
(c) ![emf(t=0.050s)=(50)(1.80T)(0.200m*0.100m)(24rad/s)sin(24.0rad/s(0.050s))\\\\emf(t=0.050s)=40.26V](https://tex.z-dn.net/?f=emf%28t%3D0.050s%29%3D%2850%29%281.80T%29%280.200m%2A0.100m%29%2824rad%2Fs%29sin%2824.0rad%2Fs%280.050s%29%29%5C%5C%5C%5Cemf%28t%3D0.050s%29%3D40.26V)
(d) The torque is given by:
![\tau=NABIsin\theta\\\\NAB\omega=emf_{max}\\\\\tau=\frac{emf_{max}}{\omega}\frac{emf_{max}}{R}\\\\\tau=\frac{(43.20V)^2}{(24.0rad/s)(10.0\Omega)}=7.77Nm](https://tex.z-dn.net/?f=%5Ctau%3DNABIsin%5Ctheta%5C%5C%5C%5CNAB%5Comega%3Demf_%7Bmax%7D%5C%5C%5C%5C%5Ctau%3D%5Cfrac%7Bemf_%7Bmax%7D%7D%7B%5Comega%7D%5Cfrac%7Bemf_%7Bmax%7D%7D%7BR%7D%5C%5C%5C%5C%5Ctau%3D%5Cfrac%7B%2843.20V%29%5E2%7D%7B%2824.0rad%2Fs%29%2810.0%5COmega%29%7D%3D7.77Nm)
Answer:
This can easily be solved using the "range" formula
R = V0^2 * sin 2θ / g
V0^2 = g R / sin 2θ = 9.80 * 43 / sin 144 = 717 m^2/s^2
V0 = 26.8 m/s