A 0.3 g mosquito is flying toward a girl with a speed of 4.5 mph. Just before landing on the girl, the fly is swatted straight back at a speed of 12 mph. If the fly swatter and the fly were in contact for 0.2 s, what is the force that was exerted on the fly

The required magnitude of the force exerted on the fly is of .

Given data:

The mass of mosquito is,

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,

Now as per the Newton's second law,

Thus, the required magnitude of the force exerted on the fly is of .

Learn more about the Newton's second law here:

link

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

your answer is c! hope this : )