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Anastaziya [24]

3 years ago

14

A Cessna 150 aircraft has a lift-off speed of approximately 125 km/h. What minimum constant acceleration does this require if th

e aircraft is to be airborne after a take-off run of 165 m? Answer in units of m/s 2 . 010 (part 2 of 3) 10.0 points What is the corresponding take-off time? Answer in units of s. 011 (part 3 of 3) 10.0 points If the aircraft continues to accelerate at this rate, what speed will it reach 15.1 s after it begins to roll? Answer in units of m/s.

1 answer:

aivan3 [116]3 years ago

7 0

Answer:

Part a)

$a = 3.65 m/s^2$

Part b)

$t = 9.5 s$

Part c)

$v_f = 55.1 m/s$

Explanation:

Part a)

As we know that it starts from rest and moves on runway by total distance 165 m

so we will have

$v_f^2 - v_i^2 = 2ad$

$v_f^2 - 0 = 2(a)(165)$

$v_f = 125 km/h = 34.7 m/s$

now we have

$a = 3.65 m/s^2$

Part b)

Now for take off time we will have

$v_f - v_i = at$

$34.7 - 0 = 3.65 t$

$t = 9.5 s$

Part c)

$v_f = v_i + at$

$v_f = 0 + (3.65)(15.1)$

$v_f = 55.1 m/s$

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One object is at rest, and another is moving. The two collide in a one-dimensional, completely inelastic collision. In other wor

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