Question

A jet plane traveling 1840 km/h (511 m/s ) pulls out of a dive by moving in an arc of radius 5.40 km .

Answer 1

Explanation:

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Answer 2

The acceleration of plane due to gravity is 4.835 m/s²

Explanation:

Given as :

The velocity of the jet plane = 1840 km/h

i.e The velocity of the jet plane =  $\dfrac{1840\times 1000}{3600}$  = v = 511 m/s

The radius of arc = r = 5.40 km = 5.40 × 1000 = 5400 m

Let the acceleration of jet = a m/s²

The value of g = 10 m/s²

According to question

∵  The plane is moving in the circular path , then centripetal acceleration will act on it

i.e   Acceleration = $\dfrac{velocity^{2} }{radius of path}$

Or,                      a = $\dfrac{v^{2} }{r}$    m/s²

Or,                      a = $\dfrac{(511 m/s )^{2} }{5400 m}$

∴                         a = 48.35 m/s²

So, The acceleration of jet = a = 48.35 m/s²

Again

The acceleration due to gravity =  $\dfrac{a}{g}$

i.e                                                 = $\dfrac{48.35}{10}$

Or,                                                = 4.835 m/s²

So, The acceleration due to gravity =  $\dfrac{a}{g}$ = 4.835 m/s²

Hence, The acceleration of plane due to gravity is 4.835 m/s² Answer