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kap26 [50]
1 year ago
13

3. [-/2 Points]

Physics
1 answer:
natita [175]1 year ago
8 0

(a) The average velocity of the car in m/s for the first leg of the run is 36.2 m/s.

(b) The average velocity (in m/s) for the total trip is 0.

<h3>Average velocity </h3>

The average velocity of the car in m/s for the first leg of the run is calculated as follows;

Average velocity = total displacement/total time

Average velocity = (760) / (21) = 36.2 M/S

<h3>Average velocity for total trip</h3>

The average velocity (in m/s) for the total trip is calculated as follows;

v =  total displacement/total time

v = 0/(26 + 21)

v = 0

Thus, the average velocity of the car in m/s for the first leg of the run is 36.2 m/s.

The average velocity (in m/s) for the total trip is 0.

Learn more about average velocity here: brainly.com/question/6504879

#SPJ1

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Answer:

15.065ft

Explanation:

To solve this problem it is necessary to consider the aerodynamic concepts related to the Drag Force.

By definition the drag force is expressed as:

F_D = -\frac{1}{2}\rho V^2 C_d A

Where

\rho is the density of the flow

V = Velocity

C_d= Drag coefficient

A = Area

For a Car is defined the drag coefficient as 0.3, while the density of air in normal conditions is 1.21kg/m^3

For second Newton's Law the Force is also defined as,

F=ma=m\frac{dV}{dt}

Equating both equations we have:

m\frac{dV}{dt}=-\frac{1}{2}\rho V^2 C_d A

m(dV)=-\frac{1}{2}\rho C_d A (dt)

\frac{1}{V^2 }(dV)=-\frac{1}{2m}\rho C_d A (dt)

Integrating

\int \frac{1}{V^2 }(dV)= - \int\frac{1}{2m}\rho C_d A (dt)

-\frac{1}{V}\big|^{V_f}_{V_i}=\frac{1}{2m}(\rho)C_d (\pi r^2) \Delta t

Here,

V_f = 60mph = 26.82m/s

V_i = 120.7m/s

m= 1600lbf = 725.747Kg

\rho = 1.21 kg/m^3

C_d = 0.3

\Delta t=7s

Replacing:

\frac{-1}{26.82}+\frac{1}{120.7} = \frac{1}{2(725.747)}(1.21)(0.3)(\pi r^2) (7)

-0.029 = -5.4997r^2

r = 2.2963m

d= r*2 = 4.592m \approx 15.065ft

4 0
3 years ago
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