The voltage across an inductor ' L ' is
V = L · dI/dt .
I(t) = I(max) sin(ωt)
dI/dt = I(max) ω cos(ωt)
V = L · ω · I(max) cos(ωt)
L = 1.34 x 10⁻² H
ω = 2π · 60 = 377 /sec
I(max) = 4.80 A
V = L · ω · I(max) cos(ωt)
V = (1.34 x 10⁻² H) · (377 / sec) · (4.8 A) · cos(377 t)
<em>V = 24.25 cos(377 t)</em>
V is an AC voltage with peak value of 24.25 volts and frequency = 60 Hz.
Answer:
The answer is the principal Quantum number (n)
Explanation:
The principal quantum number is one of the four quantum numbers associated with an atom.
It is denoted by a number n=1,2,3,4 etc
It tells both size (directly) and energy (indirectly) of an orbital.
When n=1 means it is the closest to the nucleus and is the smallest orbital and with increase in principal quantum number, it depicts that size of the orbital is increasing.
It tells the energy of the orbital as well as smaller number means less distance from nucleus and having less energy. Since electrons requires to absorb energy to jump into higher orbitals making n=2,3,4 etc. Thus electrons in the orbitals with higher n number indicates higher energy orbitals.
It's highly reactive and contains only one valence electron
Answer:
3.5m/s^2
Explanation:
From Newton's second Law of Motion
F = ma
Where F is the applied force, m is the mass of the object and a is the acceleration.
F = 350 N
Mass = 100kg
350N = 100×a
a = 350/100
a = 3.5m/s^2
The acceleration of the object will be 3.5m/s^2
