The drag force pushes opposite your motion as you ride a bicycle. If you double your speed, what happens to the magnitude of the drag force? a. The drag force goes up by a factor of 4 b. The drag force stays the same. c. The drag force decreases. d. The drag force doubles as well

F_d,1 / F_d,o =  4 , Option A

Explanation:

Given:

- The velocity of the bike V_o is doubled to 2*V_o

Find:

what happens to the magnitude of the drag force?

Solution:

- The expression of the Drag force as a function of objects area, coefficient of drag, velocity and density of medium is given by:

                                 F_d = 0.5*C_d*A*p*V^2 / 2

- In our case the force of drag with speed V_o is:

                                F_d,o = 0.5*C_d*A*p*V_o^2 / 2

- force of drag with speed 2V_o is:

                                 F_d,1 = 2*C_d*A*p*V_o^2 / 2

- Take the ratio of F_d,1 to F_d,o:

                  F_d,1 / F_d,o =  (2*C_d*A*p*V_o^2 / 2) / (0.5*C_d*A*p*V_o^2 / 2)

We get:                      F_d,1 / F_d,o =  4

Hence, by doubling the speed the magnitude of drag force is increased by a factor of 4.    

from

explanation:

wind is defined by the direction it blows from.   for example, a wind blowing from north to south is called a northerly wind.

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft