Answer:
1.1x10^-2N
Explanation:
We have the change in momentum as
P = 0.3(4.5+12)g.mph
= 0.3x0.447x(4.5+12)x10^-3
Then the force that is exerted will be
F = p/∆t
∆t = 0.2
= 0.3x0.447x(4.5+12)x10^-3/0.2
= 0.1341x16.5x10^-3/0.2
= 1.1x10^-2
Therefore the force that was exerted is equal to 1.1x10^-2
The required magnitude of the force exerted on the fly is of [tex]5.025 \times 10^{-3} \;\rm N[/tex].
Given data:
The mass of mosquito is, [tex]m =0.3 \;\rm g =3 \times 10^{-4} \;\rm kg[/tex]
The speed of flying is, u = 4.5 mph = 4.5 ( 0.447) = 2.01 m/s.
The swatting speed of mosquito is, v = 12 mph = 12 (0.447 ) = 5.36 m/s.
The time of contact is, t = 0.2 s.
In this problem, we will first calculate the change in momentum, and the change in momentum is given as,
p = m ( v - u)
Solving as,
[tex]p = 3 \times 10^{-4} (5.36 - 2.01)\\\\p = 1.005 \times 10^{-3} \;\rm kg.m/s[/tex]
Now as per the Newton's second law,
[tex]F = p/t\\\\F = 1.005 \times 10^{-3} / 0.2\\\\F= 5.025 \times 10^{-3} \;\rm N[/tex]
Thus, the required magnitude of the force exerted on the fly is of [tex]5.025 \times 10^{-3} \;\rm N[/tex].
Learn more about the Newton's second law here:
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