Answer:
The magnitude of force must you apply to hold the platform in this position = 888.89 N
Explanation:
Given that :
Workdone (W) = 80.0 J
length x = 0.180 m
The equation for this work done by the spring is expressed as:
Making the spring constant the subject of the formula; we have:
Substituting our given values, we have:
The magnitude of the force that must be apply to the hold platform in this position is given by the formula :
F = 888.89 N
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
=
= 23.04/10.3
= 2.23 m/s²
It continues to fly but now with some tangential acceleration
= 0.63 m/s²
therefore the net value of acceleration is given by the resultant of the centripetal acceleration and the tangential acceleration
so
=
=
= 2.316 m/s²
So the magnitude of net acceleration will become 2.316 m/s².
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