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3 years ago

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Terminal velocity. A rider on a bike with the combined mass of 100kg attains a terminal speed of 15m/s on a 12% slope. Assuming

that the only forces affecting the speed are the weight and the drag, calculate the drag coefficient. The frontal area is 0.9m2 .

1 answer:

Firlakuza [10]3 years ago

8 0

Answer:

0.9378

Explanation:

Weight (W) of the rider = 100 kg;

since 1 kg = 9.8067 N

100 kg will be = 980.67 N

W = 980.67 N

At the slope of 12%, the angle θ is calculated as:

$tan \ \theta = \dfrac{12}{100} \\ \\ tan \ \theta = 0.12 \\ \\ \theta = tan^{-1}(0.12) \\\\ \theta = 6.84^0$

The drag force D = Wsinθ

$\dfrac{1}{2}C_v \rho AV^2 = W sin \theta$

where;

$\rho = 1.23 \ kg/m^3$

A = 0.9 m²

V = 15 m/s

∴

Drag coefficient $C_D = \dfrac{2 *W*sin \theta}{\rho *A *V^2}$

$C_D =\dfrac{2 *980.67*sin 6.84}{1.23 *0.9 *15^2}$

$C_D =0.9378$

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4 years ago

Doss [256]

Answer:

The resultant velocity is <u>169.71 km/h at angle of 45° measured clockwise with the x-axis</u> or the east-west line.

Explanation:

Considering west direction along negative x-axis and north direction along positive y-axis

Given:

The car travels at a speed of 120 km/h in the west direction.

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Now, considering the given directions, the velocities are given as:

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Now, since $v_1\ and\ v_2$ are perpendicular to each other, their resultant magnitude is given as:

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Plug in the given values and solve for the magnitude of the resultant.This gives,

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Let the angle made by the resultant be 'x' degree with the east-west line or the x-axis.

So, the direction is given as:

$x=\tan^{-1}(\frac{|v_2|}{|v_1|})\\\\x=\tan^{-1}(\frac{120}{-120})=\tan^{-1}(-1)=-45\ deg(clockwise\ angle\ with\ the\ x-axis)$

Therefore, the resultant velocity is 169.71 km/h at angle of 45° measured clockwise with the x-axis or the east-west line.

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