Explanation:
When one coulomb charge passes through any cross section of the wire per second,the current passing is one ampere. Charge of electron ,e=1.6X10^-19C. n=1/(1.6X10^-19)=6.25X10^18.Sep 17, 2017
Answer:
Both of them.
Explanation:
They are both because when your analyzing data , that is what happen's.
Answer:
C. Cables transmit infrared waves over longer distances.
Explanation:
A.pex
Quasi frequency = 4√6
Quasi period = π√6/12
t ≈ 0.4045
<u>Explanation:</u>
Given:
Mass, m = 20g
τ = 400 dyn.s/cm
k = 3920
u(0) = 2
u'(0) = 0
General differential equation:
mu" + τu' + ku = 0
Replacing the variables with the known value:
20u" + 400u' + 3920u = 0
Divide each side by 20
u" + 20u' + 196u = 0
Determining the characteristic equation by replacing y" with r², y' with r and y with 1 in the differential equation.
r² + 20r + 196 = 0
Determining the roots:

r = -10 ± 4√6i
The general solution for two complex roots are:
y = c₁ eᵃt cosbt + c₂ eᵃt sinbt
with a the real part of the roots and b be the imaginary part of the roots.
Since, a = -10 and b = 4√6
u(t) = c₁e⁻¹⁰^t cos 4√6t + c₂e⁻¹⁰^t sin 4√6t
u(0) = 2
u'(0) = 0
(b)
Quasi frequency:
μ = 

(c)
Quasi period:
T = 2π / μ

(d)
|u(t)| < 0.05 cm
u(t) = |2e⁻¹⁰^t cos 4√6t + 5√6/6 e⁻¹⁰^t sin 4√6t < 0.05
solving for t:
τ = t ≈ 0.4045
First, let us calculate for the volume of the block of
lead using the formula:
V = l * w * h
But we have to convert all units in terms of cm:
l = 2.0 dm = 20 cm
w = 8 cm
h = 3.5 cm
Therefore the volume is:
V = (20 cm) * (8 cm) * (3.5 cm)
V = 560 cm^3
Next we convert the mass in terms of g:
m = 6.356 kg = 6356 g
Density is mass over volume, so:
density = 6356 g / 560 cm^3
density = 11.35 g / cm^3 (ANSWER)