An aircraft executes a horizontal loop with a speed of $150$ m/s with its wings banked at an angle of $12^{o}$ . The radius of the loop is $(g = 10 m / s^{2})$

10.6 km

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9.6 km

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7.4 km

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5.8 km

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Solution

The correct option is A 10.6 km

Using the relation for the radius (r) of loop.

$tan θ = \frac{v^{2}}{r g}$

or, $tan 12^{o} = \frac{{(150)}^{2}}{r \times 10}$

or, $r = \frac{2250}{0.2125} = 10.6 \times 10^{3} m = 10.6 k m$

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An aircraft executes a horizontal loop with a speed of 150 m/s with its, wings banked at an angle of $120^{\circ}$ . The radius of the loop is $(g = 10 m / s^{2})$

Q. An aircraft executes a horizontal loop with a speed of

150 m/s

with it's wings banked at an angle of

12^{\circ}

. The radius of the loop is

(g = 10 {m/s}^{2}, tan 12^{\circ} = 0.21)

Q. An aircraft executes a horizontal loop with a speed of

150 m/s

with it's wings banked at an angle of

12^{\circ}

. The radius of the loop is

(g = 10 {m/s}^{2}, tan 12^{\circ} = 0.21)

Q. An aircraft executes a horizontal loop at a speed of

720 k m / h

with its wings banked at

15^{\circ}

. What is the radius of the loop? (take

t a n 15^{\circ} = 0.27

) and

g = 10 m s^{- 2}

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An aircraft executes a horizontal loop with a speed of 150 m/s with its wings banked at an angle of 12o. The radius of the loop is (g=10m/s2)

The correct option is A 10.6 km Using the relation for the radius (r) of loop. tanθ=v2rg or, tan12o=(150)2r×10 or, r=22500.2125=10.6×103m=10.6km

An aircraft executes a horizontal loop with a speed of $150$ m/s with its wings banked at an angle of $12^{o}$ . The radius of the loop is $(g = 10 m / s^{2})$

The correct option is A 10.6 km

Using the relation for the radius (r) of loop.

$tan θ = \frac{v^{2}}{r g}$

or, $tan 12^{o} = \frac{{(150)}^{2}}{r \times 10}$

or, $r = \frac{2250}{0.2125} = 10.6 \times 10^{3} m = 10.6 k m$