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Marizza181 [45]
2 years ago
5

An airliner of mass 1.70×105kg1.70×105kg lands at a speed of 75.0 m/sm/s. As it travels along the runway, the combined effects o

f air resistance, friction from the tires, and reverse thrust from the engines produce a constant force of 2.90×105N2.90×105N opposite to the airliner's motion. What distance along the runway does the airliner travel before coming to a halt?
Physics
1 answer:
Karo-lina-s [1.5K]2 years ago
4 0

Answer:

The airliner travels 1.65 km along the runway before coming to a halt.

Explanation:

Given

Resistive forces = (2.90 × 10⁵) N = 290000 N

Mass of the airliner = (1.70 × 10⁵) kg = 170000 kg

Velocity of airliner = 75 m/s

Let the distance over moved by the airliner be equal to d

According to the work-energy theorem, the work done by the resistive forces in stopping the airliner is equal to the travelling kinetic energy of the airliner.

Work done by the resistive forces = (290000) × d = (290,000d) J

Kinetic energy of the airliner = (1/2)(170000)(75²) = 478,125,000 J

290000d = 478,125,000

d = (478,125,000/290,000)

d = 1648.7 m = 1.65 km

Hope this helps!!!

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A sailboat starts from rest and accelerates at a rate of 0.21 m/s^2 over a distance of 280 m. find the magnitude of the boat's f
sasho [114]

We use the kinematic equations,

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S= ut + \frac{1}{2} at^2                  (B)

Here, u is initial velocity, v is final velocity, a is acceleration and t is time.

Given,  u=0, a=0.21 \ m/s^2 and s= 280 m.

Substituting these values in equation (B), we get

280 \ m = 0 +\frac{1}{2} (0.21 m/s^2) t^2 \\\\ t^2 = \frac{280 \times 2}{0.21 } \\\\ t= 51.63 \ s.

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8 0
1 year ago
A particle has a velocity of v→(t)=5.0ti^+t2j^−2.0t3k^m/s.
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Answer:

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b)a=\dfrac{1}{24.83}(5i+4j-24k)\ m/s^2

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We know that acceleration a is given as

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Therefore the acceleration function a will be

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The magnitude of the acceleration will be

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The direction of the acceleration a is given as

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a)a=5 i+2t j - 6\ t^2k

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