Answer:
launch- The first stage is ignited at launch and burns through the powered ascent until its propellants are exhausted. The first stage engine is then extinguished, the second stage separates from the first stage, and the second stage engine is ignited. The payload is carried atop the second stage into orbit
powered ascent-The first stage is ignited at launch and burns through the powered ascent until its propellants are exhausted. The first stage engine is then extinguished, the second stage separates from the first stage, and the second stage engine is ignited. The payload is carried atop the second stage into orbit
coasting flight-
When the rocket runs out of fuel, it enters a coasting flight. The vehicle slows down under the action of the weight and drag since there is no longer any thrust present. The rocket eventually reaches some maximum altitude which you can measure using some simple length and angle measurements and trigonometry.
ejection charge-At the end of the delay charge, an ejection charge is ignited which pressurizes the body tube, blows the nose cap off, and deploys the parachute. The rocket then begins a slow descent under parachute to a recovery. The forces at work here are the weight of the vehicle and the drag of the parachute.
slow decent- slow downs (i guess)
recovery-A recovery period is typically characterized by abnormally high levels of growth in real gross domestic product, employment, corporate profits, and other indicators. This is a turning point from contraction to expansion and often results in an increase in consumer confidence
Explanation:
Answer:
Multiplying impulse response by t ( option D )
Explanation:
We can obtain The impulse response of strength 1 considering a unit step response by Multiplying impulse response by t .
When we consider the Laplace Domain, and the relationship between unit step and impulse, we can deduce that the Impulse response will take the inverse Laplace transform of the function ( transfer ) . Hence Multiplying impulse response by t will be used .
Answer:
Estimated number of indigenous faults remaining undetected is 6
Explanation:
The maximum likelihood estimate of indigenous faults is given by,
here,
= the number of unseeded faults = 6
= number of seeded faults = 30
= number of seeded faults found = 15
So NF will be calculated as,
And the estimate of faults remaining is 
= 12 - 6 = 6
