Answer:
Therefore the answer is the precision in the speed DECREASES
Explanation:
In quantum mechanics, we have the uncertainty principle that establishes that when the accuracy of the position increases the accuracy the speed decreases, being related by the expression
Δx Δv ≥ h'/ 2
h' = h/2π
Therefore the answer is the precision in the speed DECREASES
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°
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A. reactants
B. subscript
C. coefficient
D. products
The distance an object falls, from rest, in gravity is
D = (1/2) (G) (T²)
'T' is the number seconds it falls.
In this problem,
0.92 meter = (1/2) (9.8) (T²)
Divide each side by 4.9 : 0.92 / 4.9 = T²
Take the square root
of each side: √(0.92/4.9) = T
0.433 sec = T
The horizontal speed doesn't make a bit of difference in
how long it takes to reach the floor. BUT ... if you want to
know how far from the table the pencil lands, you can find
that with the horizontal speed.
The pencil is in the air for 0.433 second.
In that time, it travels
(0.433s) x (1.4 m/s) = 0.606 meter
from the edge of the table.
