A 650 kg airplane must reach its takeoff speed while traveling down a 61.0 m long runway. The pilot starts the plane from rest, and by the end of the runway he has reached the takeoff speed of 29.0 m/s. To the nearest newton, what net force must act on the plane? Hint: there are multiple ways you could go about solving this problem.

Respuesta :

Answer:

Force in the plane will be 4488.09 N

Explanation:

We have given mass of the airplane m = 650 kg

Length of the runway S = 61 m

As the plane starts from rest so initial; velocity u = 0 m/sec

Final velocity of airplane v = 29 m/sec

From first equation of motion v = u+at

29 = 0+at

at = 29

From second equation of motion [tex]s=ut+\frac{1}{2}at^2[/tex]

[tex]61=0\times t+\frac{1}{2}at^2[/tex]

[tex]122=at^2[/tex]

Now using at =29

[tex]122=29t[/tex]

t = 4.20 sec

So [tex]4.2a=29[/tex]

[tex]a=6.9m/sec^2[/tex]

According to second law of motion F= ma

So force F = 650×6.9 = 4488.09 N

Answer:

Explanati2/7 = 8/28

1/4 = 7/28

Therefore Mr Pham mowed more.

8+7 = 15 so 15/28 of lawn has been mowed, therefore 13/28 of the lawn is left still to be mowe.

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