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Bad White [126]
2 years ago
10

A plane accelerates from rest at a constant rate of 5.00 m/s2 along a runway that is 1800 m long. Assume that the plane reaches

the required takeoff velocity at the end of the runway. What is the time tTO needed to take off? Express your answer in seconds using three significant figures.
Physics
1 answer:
tiny-mole [99]2 years ago
3 0

Answer:

26.8 seconds

Explanation:

To solve this problem we have to use 2 kinematics equations: *I can't use subscripts for some reason on here so I am going to use these variables:

v = final velocity

z = initial velocity

x = distance

t = time

a = acceleration

{v}^{2}  =  {z}^{2}  + 2ax

v = z + at

First let's find the final velocity the plane will have at the end of the runway using the first equation:

{v}^{2}  =  {0}^{2}  + 2(5)(1800)

v = 60 \sqrt{5}

Now we can plug this into the second equation to find t:

60 \sqrt{5}  = 0 + 5t

t = 12 \sqrt{5}

Then using 3 significant figures we round to 26.8 seconds

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Explanation:

Given that,

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We need to find the angular momentum of the skater. The formula for the angular momentum of the skater is given by :

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Substitute all the values,

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At which point does the planet have the least gravitational force acting on it?
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At which point does the planet have the least gravitational force acting on it?

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Explanation:

The given data is as follows.

            m = 5000 kg,            h = 800 km = 0.8 \times 10^{6} m

    R_{e} = 6.37 \times 10^{6} m,    r = R + h = 7.17 \times 10^{6} m

   M_{e} = 5.98 \times 10^{24} kg,   G = 6.67 \times 10^{-11} Nm^{2}/kg^{2}

As we know that,

              \frac{mv^{2}}{r} = \frac{GmM_{e}}{r^{2}}

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And, it is known that formula to calculate angular velocity is as follows.

               \omega = \frac{v}{r} = \sqrt{\frac{GM_{e}}{r^{3}}}

                            v = \sqrt{\frac{GM_{e}}{r^{3}}}

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Thus, we can conclude that speed of the satellite is 1.0402 \times 10^{-3} rad/s.

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