What is meant in astronomy by the phrase "adaptive optics?

Answer:

a)

b) the distance the motorcycle covers is 155 m

Explanation:

Let denote the variables. Next, we analyze the motion equation for the accelerating motorcycle alongside the constant speed of the car:

where:

represents the motorcycle's speed at time 2

is the steady velocity of the car

indicates the initial speeds of both vehicle types at time 1

d signifies the distance separating the car and motorcycle at the initial moment

x is the distance the car travels from time 1 to time 2

Solving the equations provides:

For the second query, we determine x+d by applying the car’s motion equation to compute x:

To calculate the resultant vector, we need to consider the component vectors, which are aligned with the x and y axes. The resultant vector v is determined using the formula: v = √(vx² + vy²). Substituting the values gives us: v = √[(9.80)² + (-6.40)²]. This results in v = √137, which approximates to 11.7 units.

Quasi frequency = 4√6Quasi period = π√6/12t ≈ 0.4045Explanation: The given data is: Mass, m = 20gτ = 400 dyn.s/cmk = 3920u(0) = 2u'(0) = 0General differential equation:mu" + τu' + ku = 0Substituting the known values yields:20u" + 400u' + 3920u = 0Now, dividing each side by 20 gives:u" + 20u' + 196u = 0To establish the characteristic equation, replace y" with r², y' with r, and y with 1 in the differential equation:r² + 20r + 196 = 0Now, solving for the roots results in:r = -10 ± 4√6iThe general solution for two complex roots is: y = c₁ eᵃt cosbt + c₂ eᵃt sinbtwith a being the real part of the roots and b representing the imaginary part. Given that a = -10 and b = 4√6,u(t) = c₁e⁻¹⁰^t cos 4√6t + c₂e⁻¹⁰^t sin 4√6twhere u(0) = 2 and u'(0) = 0(b)Quasi frequency: μ = (c)(d)Quasi period: T = 2π / μ(d)|u(t)| < 0.05 cmu(t) = |2e⁻¹⁰^t cos 4√6t + 5√6/6 e⁻¹⁰^t sin 4√6t| < 0.05solving for t:τ = t ≈ 0.4045

halved

Answer:

v = 1176.23 m/s

y = 741192.997 m = 741.19 km

Explanation:

Given:

M₀ = 9 Kg  (Starting mass)

me = 0.225 Kg/s   (Fuel consumption rate)

ve = 1980 m/s    (Exhaust speed in relation to rocket, at atmospheric conditions)

v =? for t = 20 s

y =?

We apply the formula:

v = ∫((ve*me)/(M₀ - me*t)) dt - ∫g dt     where t ranges from 0 to the specified t

⇒   v = - ve*Ln ((M₀ - me*t)/M₀) - g*t

This leads us to:

v = - 1980 m/s*Ln ((9 Kg - 0.225 Kg/s*20 s)/(9 Kg)) - (9.81 m/s²)(20 s)

Thus, we find:

v = 1176.23 m/s

Next, we utilize the next formula:

y = ∫v dt = ∫(- ve*Ln ((M₀ - me*t)/M₀) - g*t) dt

⇒   y = - ve* ∫ Ln ((M₀ - me*t)/M₀) dt - g*∫t dt

⇒   y = - ve*(Ln((M₀ - me*t)/M₀)*t + (M₀/me)*(M₀  - me*t - M₀*Ln(M₀ - me*t))) - (g*t²/2)

For t = 20 s, we calculate:

y = Ln((9 Kg - 0.225 Kg/s*20 s)/9 Kg)*(20 s) + (9 Kg/0.225 Kg/s)*(9 Kg  - 0.225 Kg/s*20 s - 9 Kg*Ln(9 Kg - 0.225 Kg/s*20 s)) - (9.81 m/s²*(20 s)²/2)

⇒   y = 741192.997 m = 741.19 km

The graphs are presented in the images.