Answer:
velocity = 1527.52 ft/s
Acceleration = 80.13 ft/s²
Explanation:
We are given;
Radius of rotation; r = 32,700 ft
Radial acceleration; a_r = r¨ = 85 ft/s²
Angular velocity; ω = θ˙˙ = 0.019 rad/s
Also, angle θ reaches 66°
So, velocity of the rocket for the given position will be;
v = rθ˙˙/cos θ
so, v = 32700 × 0.019/ cos 66
v = 1527.52 ft/s
Acceleration is given by the formula ;
a = a_r/sinθ
For the given position,
a_r = r¨ - r(θ˙˙)²
Thus,
a = (r¨ - r(θ˙˙)²)/sinθ
Plugging in the relevant values, we obtain;
a = (85 - 32700(0.019)²)/sin66
a = (85 - 11.8047)/0.9135
a = 80.13 ft/s²
Answer:
E = 10t^2e^-10t Joules
Explanation:
Given that the current through a 0.2-H inductor is i(t) = 10te–5t A.
The energy E stored in the inductor can be expressed as
E = 1/2Ll^2
Substitutes the inductor L and the current I into the formula
E = 1/2 × 0.2 × ( 10te^-5t )^2
E = 0.1 × 100t^2e^-10t
E = 10t^2e^-10t Joules
Therefore, the energy stored in the inductor is 10t^2e^-10t Joules
Complementary angles are those whose sum is 90 degrees, right?, then let's find A, A=72 degrees, because its mid point is 36 deg. Then, B = 18 degrees and its mid point is 9 degrees! :)
