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gayaneshka [121]
3 years ago
11

Suppose a plane accelerates from rest for 30 s, achieving a takeoff speed of 80 m/s after traveling a distance of 1200 m down th

e runway. A smaller plane with the same acceleration has a takeoff speed of 40 m/s. Starting from rest, after what distance will this smaller plane reach its takeoff speed?
Physics
1 answer:
Margaret [11]3 years ago
5 0

Answer:

300 m

Explanation:

The train accelerate from the rest so u = 0 m/sec

Final speed that is v = 80 m/sec

Time t = 30 sec

The distance traveled by first plane = 1200 m

We know the equation of motion S=ut+\frac{1}{2}at^2 where s is distance a is acceleration and u is initial velocity

Using this equation for first plane 1200=0\times 30+\frac{1}{2}a30^2

a=2.67\frac{m}{sec^2}

As the acceleration is same for both the plane so a for second plane will be 2.67 \frac{m}{sec^2}

The another equation of motion is v^2=u^2+2as using this equation for second plane 40^2=0+2\times 2.67\times s

s = 300 m

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Answer:

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From the question we are told that

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=>   E  A  =  \frac{q}{\epsilon_o }

Given that  the electric field is pointing downward  , the equation become

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Here   q is the excess charge on the surface of the earth

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