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victus00 [196]
3 years ago
12

A jetliner, traveling northward, is landing with a speed of 69 m/s. Once the jet touches down, it has 750 m of runway in which t

o reduce its speed to 6.1 m/s. Compute the average acceleration (magnitude and direction) of the plane during landing.
Physics
1 answer:
Katena32 [7]3 years ago
3 0

Answer:

Acceleration, a=-3.14\ m/s^2

Explanation:

It is given that,

Initial speed of the jetliner, u = 69 m/s

Final speed of the jetliner, v = 6.1 m/s

Distance covered, s = 750 m

We need to find the average acceleration of the jetliner. It can be calculated using third equation of motion as :

a=\dfrac{v^2-u^2}{2s}

a=\dfrac{(6.1)^2-(69)^2}{2\times 750}          

a=-3.14\ m/s^2

So, the average acceleration of the plane during landing is -3.14\ m/s^2. Hence, this is the required solution.

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Answer:

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Explanation:

The final velocity of the raindrop as it reaches the ground can be found by using the equation for a uniformly accelerated motion:

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v=\sqrt{u^2 +2gd}=\sqrt{0^2+2(9.8 m/s^2)(2000)}=198 m/s

And keeping in mind that

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The speed converted into miles per hour is

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Answer:

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The mechanical energy of a body or a physical system is the sum of its kinetic energy and the potential energy. According to the Principle of Conservation of Energy for mechanical energy, the total mechanical energy that a body possesses is constant at every instant of time.

Since mechanical energy is equal to the sum of kinetic energy and gravitational potential energy that a body possesses, the only way to stay constant is that:

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Due to the Principle of Conservation of Energy you can say that the gravitational potential energy is converted to kinetic energy.  So Gravitational potential energy at the top = kinetic energy at the bottom

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