Answer:
i. 43.5 mH ii. 16 Ω. In phasor form Z = (8.33 + j13.66) Ω iii 58.64°
Explanation:
i. The resistance , R of the non-inductive load R = 125 V/15 A = 8.33 Ω
The reactance X of the inductor is X = 2πfL where f = frequency = 50 Hz.
So, x = 2π(50)L = 100πL Ω = 314.16L Ω
Since the current is the same when the 240 V supply is applied, then
the impedance Z = √(R² + X²) = 240 V/15 A
√(R² + X²) = 16 Ω
8.33² + X² = 16²
69.3889 + X² = 256
X² = 256 - 69.3889
X² = 186.6111
X = √186.6111
X = 13.66 Ω
Since X = 314.16L = 13.66 Ω
L = 13.66/314.16
= 0.0435 H
= 43.5 mH
ii. Since the same current is supplied in both circuits, the impedance Z of the circuit is Z = 240 V/15 A = 16 Ω.
So in phasor form Z = (8.33 + j13.66) Ω
iii. The phase difference θ between the current and voltage is
θ = tan⁻¹X/R
= tan⁻¹(314.16L/R)
= tan⁻¹(314.16 × 0.0435 H/8.33 Ω)
= tan⁻¹(13.66/8.33)
= tan⁻¹(1.6406)
= 58.64°
<span>This question is based on conservation of energy as the work done would lead to change in kinetice energy of car
change in KE = 1/2 mv(f)^2 - 1/2mv(i)^2 = 1/2m(v(f)^2-v(i)^2)
where v(f) and v(i) are the final and initial speeds
change in KE = 185kJ = 185,000J = 1/2 m((28m/s)^2-(23m/s)^2)
185,000=1/2 m(255m^2/s^2)
solving for m
m=1451kg</span>
He arranged his periodic table by each elements atomic mass
Answer:
Lower
Lower
gsintheta (gsinθ)
Explanation:
The sum of forces resolved parallel to the inclined plane is given by;
F - mgsinθ = 0
ma - mgsinθ = 0
ma = mgsinθ
a = gsinθ
Acceleration is proportional to angle of inclination, thus the lower the angle of the slope, lower the acceleration along the ramp.
therefore, the speed at the bottom of a slope will be lower, (velocity is directly proportional to acceleration) and, consequently, the control will be better.
The acceleration along the ramp, is gsintheta (gsinθ)
