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Vikki [24]
3 years ago
14

An airplane flies eastward and always accelerates at a constant rate. at one position along its path it has a velocity of 26.3 m

/s, it then flies a further distance of 48100 m and afterwards its velocity is 41.9 m/s. find the airplane\'s acceleration and calculate how much time elapses while the airplane covers those 48100 m.
Physics
1 answer:
galina1969 [7]3 years ago
6 0
We will use the formula / equation to determined the time.

Distance = ½ * (vi + vf) * t 
48100 = ½ * (26.3 + 41.9) * t 
t = 48100 ÷ 34.1 = 1410.557185 seconds 

We will use the formula / equation to determined the acceleration. 

vf = vi + a * t 
41.9 = 26.3 + a * 1410.557185 
a = (41.9 – 26.3) ÷ 1410.557185 = 0.011059459 m/s^2 

We will use the formula / equation to determined the acceleration.

vf^2 = vi^2 + 2 * a * d 
41.9^1 = 26.3^2 + 2 * a * 48100 
a = (41.9^2 – 26.3^2) ÷ 96200 = 0. 011059459 m/s^2 
Since both answers are the same, I believe the acceleration is correct.
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Answer:

The work done to get you safely away from the test is  2.47 X 10⁴ J.

Explanation:

Given;

length of the rope, L = 70 ft

mass per unit length of the rope, μ = 2 lb/ft

your mass, W = 120 lbs

mass of the 70 ft rope  = 2 lb/ft x 70 ft

                                         = 140 lbs.

Total mass to be pulled to the helicopter, M = 120 lbs  + 140 lbs  

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The work done is calculated from work-energy theorem as follows;

W = Mgh

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h is height the total mass is raised = length of the rope = 70 ft

W = 260 Lb x 32.17 ft/s²  x 70 ft

W = 585494 lb.ft²/s²

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