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An airplane flies in a loop (a circular path in a vertical plane) of radius 160 m . The pilot's head always points toward the center of the loop. The speed of the airplane is not constant; the airplane goes slowest at the top of the loop and fastest at the bottom

(a) At the top of the loop, the pilot feels weightless. What is the speed of the airplane at this point?
(b) At the bottom of the loop, the speed of the airplane is 300. What is the apparent weight of the pilot at this point? His true weight is 760 .

Respuesta :

Answer:

a) 39.6 m/s b) 4123 N

Explanation:

a) At the top of the loop, all of the forces point downwards (force of gravity and normal force).

Fnet=ma

ma=m(v^2/R) (centripetal acceleration)

mg=m(v^2/R)

m cancels out (this is why pilot feels weightless) so,

g=(v^2/R)

9.8 m/s^2 = v^2/160 m

v^2=1568 m^2/s^2

v=39.6 m/s

b) At the bottom of the loop, the normal force and the force of gravity point in opposite directions. The normal force is the weight felt.

Convert 300 km/hr to m/s

300 km/hr=83.3 m/s

Convert pilot's weight into mass:

760 N = 77.55 kg

Fnet=ma

n-mg=m(v^2/R)

n=(77.55 kg)(((83.3 m/s)^2)/160 m)+(77.55 kg)(9.8 m/s^2)

n=3363.2 N+760 N=4123 N

a) The speed of the airplane at this point is 39.6 m/s

b) The apparent weight of the pilot at this point is 7604123 N

Calculation of the speed & weight:

a)

We know that

Fnet=ma

Here we apply centripetal acceleration

[tex]ma=m(v^2\div R)[/tex]

So,

[tex]mg=m(v^2\div R)\\\\g=(v^2\div R)9.8 m/s^2 = v^2\div 160 m\\\\v^2=1568 m^2\div s^2[/tex]

v=39.6 m/s

b)

Here we need to Convert 300 km/hr to m/s

So,

300 km/hr=83.3 m/s

Now Convert pilot's weight into mass:

760 N = 77.55 kg

Now

Fnet=ma

[tex]n-mg=m(v^2\div R)\\\\n=(77.55 kg)(((83.3 m/s)^2)/160 m)+(77.55 kg)(9.8 m/s^2)[/tex]

n=3363.2 N+760 N

=4123 N

Learn more about speed here: https://brainly.com/question/24831712

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