Answer:
3. 560W
Explanation:
First we calculate the work required for the 700N weigh student to climb up by 8 meters. Work is force times distance
W = Fs = 700 * 8 = 5600J
The average power is defined as the rate of energy expended over a unit of time. In order words, work divided by time duration
P = W/t = 5600 / 10 = 560 W
1,000 milligrams = 1 gram
2,000 milligrams = 2 grams
3,000 milligrams = 3 grams
4,000 milligrams = 4 grams
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
Answer is B since electric field lines flow from negative to positive not from positive to positive.