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garik1379 [7]
3 years ago
6

It's possible for a determined group of people to pull an aircraft. Drag is negligible at low speeds, and the only force impedin

g motion is the rolling friction of the rubber tires on the concrete runway. In 2000, a team of 60 British police officers set a world record by pulling a Boeing 747, with a mass of 200,000 kg, a distance of 100 m in 53 s. The plane started at rest. Suppose that the Rolling Friction Coefficient = 0.02.
Estimate the force with which each officer pulled on the plane, assuming constant pulling force and constant acceleration.
Physics
1 answer:
gogolik [260]3 years ago
3 0

Answer:

F_o = 890.67 N

Explanation:

initial speed of the aircraft = 0

it covers a distance = 100 m

time taken by it = 53 s

so we will have

d = v_i t + \frac{1}{2}at^2

100 = 0 + \frac{1}{2}a(53^2)

a = 0.071 m/s^2

now we know that there are two forces on the aircraft

1) applied force in forward direction

2) friction force in opposite direction

so we will have

F - \mu mg = ma

F - (0.02)(200,000)(9.81) = 200,000(0.071)

F = 39,240 + 14,200

F = 53440 N

so net force by 60 officers is 53440 N

so force due to each officer is given as

F_o = \frac{53440}{60} = 890.67 N

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Answer:

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Explanation:

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I have a combination of myopia and presbyopia—overall, the power of my visual system is too large, but I also have a very limite
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Answer:

The range of powers is    - 5 \ D \le P \le - 2.667\  D

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From the question we are told that

       The far point of the left eye is n_f = 20 cm

       The near point of the left eye is  n =  15cm

       The near point with the glasses on is n_g =25 \ cm

     

From these parameter we can see that with the glass on that for near point the

         Object distance would be u = -25 \ cm

          Image distance would be  v =  -15 \ cm

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              \frac{1}{f} =  \frac{1}{v}  -  \frac{1}{u}

substituting values

              \frac{1}{f} =  \frac{1}{-15}  -  \frac{1}{-25}

               f =  - \frac{75}{2} cm

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               f =  - \frac{75}{2} * \frac{1}{100}

               f =  - \frac{75}{200} \ m

   Generally the power of the lens is mathematically represented as

                P  = \frac{1}{f}

Substituting values

                 P = -  \frac{200}{75}  m

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From these parameter we can see that with the glass on that for far  point the

         Object distance would be u_f = - \infty \ cm

          Image distance would be  v_f =  -20  \ cm

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                    \frac{1}{f_f} =  \frac{1}{v_f}  -  \frac{1}{u_f}

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                P  = \frac{1}{f_f}

Substituting values

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                  - 5 \ D \le P \le - 2.667\  D

   

               

               

           

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