contestada

An airplane is flying in a horizontal circle at a speed of 104 m/s. The 85.0 kg pilot does not want the centripetal acceleration to exceed 6.35 times free-fall acceleration.
(a) Find the minimum radius of the plane’s path.
(b) At this radius, what is the magnitude of the net force that maintains circular motion exerted on the pilot by the seat belts, the friction against the seat, and so forth?

Respuesta :

zainsubhani

Answer:

Part A:

[tex]r\geq 173.806 m[/tex]

Part B:

[tex]F=5289.5757 N[/tex]

Explanation:

Part A:

Airplane is moving in circular path so acceleration in circular path is given by:

[tex]a_c=\frac{V^2}{r}[/tex]

Where:

V is the velocity of object in circular path

r is the radius of circular path

In order to find radius r. Above equation will become:

[tex]r=\frac{V^2}{a_c}[/tex]

V=104 m/s

a_c=6.35*g=6.35*9.8=62.23 m/s^2

[tex]r\geq \frac{104^2}{62.23}\\ r\geq 173.806 m[/tex]

Part B:

Force on a object moving in a circular path is:

[tex]F=\frac{mV^2}{r}\\[/tex]

r=173.806 m

m=85 kg

V=104 m/s

[tex]F=\frac{85*(104)^2}{173.806} \\F=5289.5757 N[/tex]

Cricetus

(a) The minimum radius will be "173.806 m".

(b) The magnitude of net force will be "5289.58 N".

Given:

  • Velocity, [tex]V = 104 \ m/s[/tex]
  • Mass, [tex]m = 85 \ kg[/tex]

(a)

We know,

→ [tex]a_c = mg[/tex]

       [tex]= 6.35\times g[/tex]

       [tex]= 6.35\times 9.8[/tex]

       [tex]= 62.23 \ m/s^2[/tex]

Now,

Acceleration in circular path will be:

→ [tex]a_c = \frac{V^2}{r}[/tex]

or,

→ [tex]r = \frac{V^2}{a_c}[/tex]

By substituting the values, we get

     [tex]= \frac{(104)^2}{62.23}[/tex]

     [tex]= 173.806 \ m[/tex] (radius)

(b)

The force on object will be:

→ [tex]F = \frac{mV^2}{r}[/tex]

      [tex]= \frac{85\times (104)^2}{173.806}[/tex]

      [tex]= 5289.58 \ N[/tex]

Thus the approach above is correct.

Learn more about net force here:

https://brainly.com/question/14473883

Otras preguntas