Yp(t) = A1 t^2 + A0 t + B0 t e(4t)
=> y ' = 2A1t + A0 + B0 [e^(4t) +4 te^(4t) ]
y ' = 2A1t + A0 + B0e^(4t) + 4B0 te^(4t)
=> y '' = 2A1 + 4B0e(4t) + 4B0 [ e^(4t) + 4te^(4t)
y '' = 2A1 + 4B0e^(4t) + 4B0e^(4t) + 16B0te^(4t)
Now substitute the values of y ' and y '' in the differential equation:
<span>y′′+αy′+βy=t+e^(4t)
</span> 2A1 + 4B0e^(4t) + 4B0e^(4t) + 16B0te^(4t) + α{2A1t + A0 + B0e^(4t) + 4B0 te^(4t) } + β{A1 t^2 + A0 t + B0 t e(4t)} = t + e^(4t)
Next, we equate coefficients
1) Constant terms of the left side = constant terms of the right side:
2A1+ 2αA0 = 0 ..... eq (1)
2) Coefficients of e^(4t) on both sides
8B0 + αB0 = 1 => B0 (8 + α) = 1 .... eq (2)
3) Coefficients on t
2αA1 + βA0 = 1 .... eq (3)
4) Coefficients on t^2
βA1 = 0 ....eq (4)
given that A1 ≠ 0 => β =0
5) terms on te^(4t)
16B0 + 4αB0 + βB0 = 0 => B0 (16 + 4α + β) = 0 ... eq (5)
Given that B0 ≠ 0 => 16 + 4α + β = 0
Use the value of β = 0 found previously
16 + 4α = 0 => α = - 16 / 4 = - 4.
Answer: α = - 4 and β = 0
Answer:
a) W = 3.87 10⁻⁴ J
, b) P = 3.10 10⁻³ Pa
, c) λ = 671. 6 nm
, d) frequency does not change
, e) Emax = 1.39 C / m and f) Bmax = 4.7 10⁻⁹ T
Explanation:
a) Let's use the concepts of power that is work for the unit of time and work is the change of kinetic energy
P = W / t
An electromagnetic wave has an intensity
I = P / A =
W = P t
W = 258 10⁻³ 1.5 10⁻³
W = 3.87 10⁻⁴ W s
W = 3.87 10⁻⁴ J
b) the radiation pressure is given by the ratio
P = I / c
Where I is the intensity
I = Powers / A
A = π r² = π (d/2)²
I = 258 10⁻³ / π (297.5 10⁻⁶)²
I = 9.29 10 5 W / m²
P = 9.29 10⁵/3 10⁸
P = 3.10 10⁻³ Pa
c) The wavelength when passing a measured of different refractive index changes in the way
λ = λ₀ / n
λ = 900 10⁻⁹ / 1.34
λ = 671.6 10⁻⁹ m
λ = 671. 6 nm
d) when the light strikes a medium creates a forced oscillation in the electrons of the medium, this is a resonance phenomenon, so the frequency does not change
e) The maximum electric field is
I = Emax2 / 2 μ₀ c
Emax = Ra (2 μ₀ c I)
Emax = Ra (2 4 π 10⁻⁷ 3 10⁸ 258 10⁻³)
Emax = 1.39 C / m
f) the elective and magnetic fields are related
c = Emax Bmax
Bmax = Emax / c
Bmax = 1.39 / 3 10⁸
Bmax = 4.7 10⁻⁹ T
Answer:
Explanation:
Given
acceleration of rocket(a)
At h=1000 m rocket burn out
(b) time to reach v=100 m/s
v=u+at
(c)Rocket maximum altitude
here u=100 m/s
v=0
[tex]s=510.204
Therefore maximum altitude=510.204+1000=1510.204 m
