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1 year ago

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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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:

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

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$\rho$ is the density of the flow

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A = Area

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

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Equating both equations we have:

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

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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:

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

$-0.029 = -5.4997r^2$

$r = 2.2963m$

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