Question

a hawk flies in a horizontal arc of radius 10.3 m at a constant speed of 4.8 m/s. find its centripetal acceleration. answer in units of m/s 2 . 013 (part 2 of 2) 10.0 points it continues to fly along the same horizontal arc but increases its speed at the rate of 0.63 m/s 2 . find the magnitude of acceleration under these new conditions. answer in units of m/s 2 .

Answer 1

The hawk’s centripetal acceleration is 2.23 m/s²

The magnitude of the acceleration under new conditions is 2.316 m/s²

radius of the horizontal arc = 10.3 m

the initial constant speed = 4.8 m/s

we know that the centripetal acceleration is given by

    $a_{c}$  = $\frac{v^{2} }{r}$

   $a_{c}$  = 23.04/10.3

    $a_{c}$  = 2.23 m/s²

It continues to fly but now with some tangential acceleration

$a_{t}$ = 0.63 m/s²

therefore the net value of acceleration is given by the resultant of the centripetal acceleration and the tangential acceleration

so

$a_{net}$  =  $\sqrt{a_{c} ^{2} +a_{t} ^{2} }$

$a_{net}$  =  $\sqrt{4.97 + 0.396}$

$a_{net}$  =  2.316 m/s²

So the magnitude of  net acceleration will become 2.316 m/s².

learn more about acceleration here :

brainly.com/question/11560829

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