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Luba_88 [7]
2 years ago
6

An airplane has a takeoff speed of 62 m/s. assuming a constant acceleration of 1.7 m/s2, what is the minimum length of runway it

needs for takeoff?
Physics
1 answer:
UkoKoshka [18]2 years ago
3 0

Data:

u=0 m/s is the initial velocity of the plane

v=62 m/s is the final velocity of the plane (at which the plane takes off)

a=1.7 m/s^2 is the acceleration of the plane

To find the minimum distance S the plane needs to take off, we can use the following equation:

2aS=v^2 -u^2

Re-arranging it and substituting the numbers, we find

S=\frac{v^2-u^2}{2a}=\frac{(62 m/s)^2-0}{2(1.7 m/s^2)}=1131 m

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Riders in an amusement park ride shaped like a Viking ship hung from a large pivot are rotated back and forth like a rigid pendu
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Answer:

a)  v = 16.57 m / s, b)  a = 19.6 m / s², d)    N = 1.76 10³ N,     N / W = 3

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This exercise looks interesting, but I think you have some problem with the writing, the questions seem a bit disconnected from the initial text.

Let's answer the questions.

a) For this part we can use energy considerations.

Starting point. The upper part of the trajectory indicates that the arm is horizontally

          Em₀ = U = m g h

in this case h = r

Final point. For lower of the trajectory

          Em_f = K = ½ m v²

as they indicate that there is no friction

         Em₀ = em_f

         mgh = ½ m v²

         v = \sqrt{2gh}

let's calculate

        v = \sqrt{2 \ 9.8 \ 14.0}

         v = 16.57 m / s

b) the centripetal acceleration has the formula

           a = v² / r

           a = 16.57² / 14.0

           a = 19.6 m / s²

c) see attached where the diagram is

where N is the normal and w the weight

d) let's use Newton's second law

               N-W = m a

               N - mg = m ar

               N = m (g + a)

let's calculate

               N = 60.0 (9.8 + 19.6)

               N = 1.76 10³ N

the relationship with weight is

              N / W = 1.76 10³/( 60 9.8)

              N / W = 3

normal is three times greater than body weight

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