For the honeybee that circles a field looking for a flower upon which to land, we have:
a) The tangential velocity is 10.01 m/s.
b) The centripetal acceleration is 40.1 m/s².
c) The centripetal force is 1.60 N.
a) The tangential velocity can be calculated as follows:
[tex] v = \omega r = \frac{2\pi r}{T} [/tex] (1)
Where:
r: is the radius of the circle = 2.5 m
ω: is the angular velocity = 2π/T
T: is the period = 1.57 s
By entering the above values into equation (1), we have:
[tex] v = \frac{2\pi r}{T} = \frac{2\pi 2.5 m}{1.57 s} = 10.01 m/s [/tex]
Hence, the tangential velocity is 10.01 m/s.
b) The centripetal acceleration is related to the tangential velocity as follows:
[tex] a_{c} = \frac{v^{2}}{r} = \frac{(10.01 m/s)^{2}}{2.5 m} = 40.1 m/s^{2} [/tex]
Therefore, the centripetal acceleration is 40.1 m/s².
c) The centripetal force is given by:
[tex] F = ma_{c} [/tex]
Where:
m: is the honeybee's mass = 0.04 kg
[tex] F = ma_{c} = 0.04 kg*40.1 m/s^{2} = 1.60 N [/tex]
Hence, the centripetal force is 1.60 N.
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