The <span>asthenosphere is under the lithosphere.</span>
Answer: 0.43 V
Explanation:
L = [μ(0) * N² * A] / l
Where
L = Inductance of the solenoid
N = the number of turns in the solenoid
A = cross sectional area of the solenoid
l = length of the solenoid
7.3*10^-3 = [4π*10^-7 * 450² * A] / 0.24
1.752*10^-3 = 4π*10^-7 * 202500 * A
1.752*10^-3 = 0.255 * A
A = 1.752*10^-3 / 0.255
A = 0.00687 m²
A = 6.87*10^-3 m²
emf = -N(ΔΦ/Δt).........1
L = N(ΔΦ/ΔI) so that,
N*ΔΦ = ΔI*L
Substituting this in eqn 1, we have
emf = - ΔI*L / Δt
emf = - [(0 - 3.2) * 7.3*10^-3] / 55*10^-3
emf = 0.0234 / 0.055
emf = 0.43 V
Answer:
(a) 6.567 * 10^15 rev/s or hertz
(b) 8.21 * 10^14 rev/s or hertz
Explanation:
Fn= 4π^2k^2e^4m * z^2/(h^3*n^3)
Where Fn is frequency at all levels of n.
Z = 1 (nucleus)
e = 1.6 * 10^-19c
m = 9.1 * 10^-31 kg
h = 6.62 * 10-34
K = 9 * 10^9 Nm2/c2
(a) for groundstate n = 1
Fn = 4 * π^2 * (9*10^9)^2*(1.6*10^-19)^4* (9.1 * 10^-31) * 1 / (6.62 * 10^-31)^3 = 6.567 * 10^15 rev/s
(b) first excited state
n = 1
We multiple the groundstate answer by 1/n^3
6.567 * 10^15 rev/s/ 2^3
F2 = 8.2 * 10^ 14 rev/s
