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Kitty [74]
3 years ago
6

An Airbus A350 is initially moving down the runway at 6.0 m/s preparing for takeoff. The pilot pulls on the throttle so that the

engines give the plane a constant acceleration of 1.8 m/s^2. The plane then travels a distance of 1500 m down the runway before lifting off. How long does it take from the application of the acceleration until the plane lifts off, becoming airborne?
Physics
1 answer:
alexdok [17]3 years ago
8 0

Answer:

t=67.7s

Explanation:

From this question we know that:

Vo = 6m/s

a = 1.8 m/s2

D = 1500m

And we also know that:

X=V_{o}*t + \frac{a*t^{2}}{2}   Replacing the known values:

1500=6t+0.9*t^{2}    Solving for t we get 2 possible answers:

t1 = -44.3s   and t2 = 67.7s    Since negative time represents an instant before the beginning of the movement, t1 is discarded. So, the final answer is:

t = 67.7s

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2 years ago
A mine car (mass=390 kg) rolls at a speed of 0.50 m/s on a horizontal track, as the drawing shows. A 250-kg chunk of coal has a
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Answer:

v=0.60 m/s

Explanation:

Given that

m ₁= 390 kg ,u ₁= 0.5 m/s

m₂ = 250 kg ,u₂ = 0.76 m/s

As we know that if there is no any external force on the system the total linear momentum of the system will be conserve.

Pi = Pf

m ₁u ₁+m₂u₂ = (m₂ + m ₁ ) v

Now putting the values in the above equation

390 x 0.5 + 250 x 0.76 = (390 + 250 ) v

v=\dfrac{390\times 0.5+250\times 0.76}{390+250}\ m/s

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8 0
3 years ago
Joe and Max shake hands and say goodbye. Joe walks east 0.40 km to a coffee shop, and Max flags a cab and rides north 3.65 km to
katen-ka-za [31]

Answer:

3.67 km

Explanation:

Joe distance towards coffee shop is,

OB=0.40 km

And the Max distance towards bookstore is,

OA=3.65 km

Now the distance between the Joy and Max will be,

By applying pythagorus theorem,

AB=\sqrt{OB^{2}+OA^{2}}

Substitute 0.40 km for OB and 3.65 km for OA in the above equation.

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Therefore the distance between there destination is 3.67 km.

6 0
3 years ago
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