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°
Answer:
Explanation:
ASSUMING that block = sled AND that the rope is parallel to the slope.
The force acting parallel due to the weight is
13.6(9.81)sin35.5 = 77.475 N
The maximum friction force is
(0.45)13.6(9.81)cos35.5 = 48.877 N
If rope tension is T
77.475 - 48.877 < T < 77.475 + 48.877
28.6 N < T < 126 N
28.6 N will occur if the block is on the verge of sliding downhill
126 N will occur if the block is on the verge of sliding uphill
Could be any value between them.
It's 3.6 meters per second less than my speed was
at 4:19 PM last Tuesday.
Does that tell you anything ?
Why not ?
Plugging in for the Earth's mass and for G, we have 11.2 km/s for the escape velocity for an object launched from the Earth's surface. This is about 25,000 miles per hour
In this item, we are asked to determine the speed of the bobsled given the distance traveled and the time it takes to cover the certain distance. This can mathematically be expressed as,
speed = distance / time
Substituting the given values in this item,
speed = (113 m) / (29 s)
speed = 3.90 m/s
<em>ANSWER: 3.90 m/s</em>
