Answer:
yes it is
Explanation:
it's an extremely good phone in my opinion
Answer:
The circuit impedance is 3.84 phase 38.65º and the voltage across the capacitor is 0.13 phase -128.65º V.
Explanation:
Since the voltage given to us was Vs = 5*cos(5t) V and it is the form of V = Vmax*cos(omega*t) V we can extract the frequency omega, wich is w = 5 rad/s.
In the circuit we have a capacitor and a inductor. The capacitor impedance is negative and it is inversely proportional to the frequency, while the inductor impedance is positive and directly proportional to the frequency. So we have:
Z = R + jw*L - j/(wC)
Z = 3 + j*5*0.5 - j/(5*2)
Z = 3 + j*2.5 - j*0.1 = 3 + j*2.4 Ohm = 3.84 phase 38.65º Ohm
To find out the voltage across the capacitor we can use a voltage divider equation that is:
Vcapacitor = [Zcapacitor/(R + Zinductor + Zcapacitor)] * Vsource
Vcapacitor = [(-j0.1)/(3 + j2.4)]*Vsource
Vcapacitor = [(0.1 phase -90º)/(3.84 phase 38.65º)]*5 phase 0º
Vcapacitor = [0.026 phase -128.65º]* 5 phase 0º
Vcapacitor = 0.13 phase -128.65º V
Given Information:
Capacitor 1 = C₁ = 100 μF
Capacitor 2 = C₂ = 300 μF
Voltage rating of capacitor 1 = 25 V
Voltage rating of capacitor 2 = 35 V
Supply voltage = Vs = 100 V
Required Information:
Which capacitor will be damaged = ?
Answer:
Capacitor 1 will get damaged
Explanation:
The given two capacitors are connected in series, the equivalent capacitance will be
The voltage across the capacitor 1 is
Since the voltage rating of capacitor 1 is 25 volts, it will get damaged because 75 volts are much greater as compared to its voltage rating.
The voltage across the capacitor 2 is
Since the voltage rating of capacitor 2 is 35 volts, it will not get damaged because 25 volts are less as compared to its voltage rating.
Answer:
MPa
T = 1536.8 N m
Explanation:
Given data:
Torque = 3.5 k N m = 3.5*10^3 N.m
Diameter D = 5 cm = 0.05 m
a) from torsional equation we have
solving for
MPa
B)
Pa
m
m
T = 1536.8 N m
Friction losses in pipes can be reduced by decreasing the length of the pipes, reducing the surface roughness of the pipes, and increasing the pipe diameter. Thus, options (c),(e), and (f) hold correct answers.
Friction loss is a measure of the amount of energy a piping system loses because flowing fluids meet resistance. As fluids flow through the pipes, they carry energy with them. Unfortunately, whenever there is resistance to the flow rate, it diverts fluids, and energy escapes. These opposing forces result in friction loss in pipes.
Friction loss in pipes can decrease the efficiency of the functions of pipes. These are a few ways by which friction loss in pipes can be reduced and the efficiency of the piping system can be boosted:
- <u><em>Decrease the length of the pipes</em></u>: By decreasing pipe lengths and avoiding the use of sharp turns, fittings, and tees, whenever possible result in a more natural path for fluids to flow.
- <u><em>Reduce the surface roughness of the pipes</em></u>: By reducing the interior surface roughness of pipes, a smooth and clearer path is provided for liquids to flow.
- <u><em>Increase the pipe diameter: </em></u>By widening the diameters of pipes, it is ensured that fluids squeeze through pipes easily.
You can learn more about friction losses at
brainly.com/question/13348561
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