Answer:
For Pentium 4 Prescott:
% of Static Power = 10
For core i5 Ivy Bridge:
% of Static Power = 43
Given Information:
Static Power of P4 = 10 W
Dynamic Power of P4 = 90 W
Static Power of i5 = 30 W
Dynamic Power of i5 = 40 W
Required Information:
% of static power w.r.t total power dissipation = ?
Explanation:
For Pentium 4 Prescott:
% of static power = static power/total power * 100
% of static power = 10/(10 + 90) * 100
% of static power = 10/(100) * 100
% of static power = 10
For core i5 Ivy Bridge:
% of static power = static power/total power * 100
% of static power = 30/(30 + 40) * 100
% of static power = 30/(70) * 100
% of static power = 43 (rounded to nearest whole integer)
Answer:
182.9 Volts
Explanation:
R = resistance of the resistor = 50 Ω
C = capacitance of the capacitor = 200 μF = 200 x 10⁻⁶ F
L = Inductance of the inductor = 120 mH = 0.12 H
f = frequency = 60 Hz
Capacitive reactance is given as
X = (2πfC)⁻¹
X = (2(3.14) (60) (200 x 10⁻⁶))⁻¹
X = 13.3 Ω
Inductive reactance is given as
X' = 2πfL
X' = 2(3.14) (60) (0.12)
X' = 45.2 Ω
Impedance of the circuit is given as
z = √(R² + (X' - X)²)
z = √(50² + (45.2 - 13.3)²)
z = 59.31 Ω
V = rms emf of the source = 240 Volts
rms voltage across the inductor is given as
V' = V z⁻¹ X'
V' = (240) (59.31)⁻¹ (45.2)
V' = 182.9 Volts
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i would choose a for 945 and a for faster
Mechanical advantage = (output force) / (input force)
MA = (2,000 N) / (120 N)
<em>MA = 16.67</em> (16 and 2/3)
