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avanturin [10]
3 years ago
6

Like the filters falling through the air, a car on the freeway represents an object with a high Reynolds number traveling throug

h a fluid. If you increase your speed from 50 mph to 75 mph, what is the percent change in the drag force that your car experiences (your fuel consumption will also roughly increase by this same percentage)?
Physics
1 answer:
Goshia [24]3 years ago
3 0

Answer:

ΔF=125.22 %

Explanation:

We know that drag force on the car given as

F_D=\dfrac{1}{2}\rho C_DA v^2

C_D=Drag coefficient

A=Projected area

v=Velocity

ρ=Density

All other quantity are constant so we can say that drag force and velocity can be given as

\dfrac{F_D_1}{F_D_2}=\dfrac{v_1^2}{v_2^2}

Now by putting the values

\dfrac{F_D_1}{F_D_2}=\dfrac{v_1^2}{v_2^2}

\dfrac{F_D_1}{F_D_2}=\dfrac{50^2}{75^2}

\dfrac{F_D_1}{F_D_2}=0.444

Percentage Change in the drag force

\Delta F=\dfrac{F_D_2-F_D_1}{F_D_1}\times 100

\Delta F=\dfrac{F_D_2-0.444F_D_2}{0.444F_D_2}\times 100

\Delta F=\dfrac{1-0.444}{0.444}\times 100

ΔF=125.22 %

Therefore force will increase by 125.22  %.

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How many revolutions per minute would a 23 m -diameter Ferris wheel need to make for the passengers to feel "weightless" at the
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Answer:

Approximately 6.2\; {\rm rpm}, assuming that the gravitational field strength is g = 9.81\; {\rm m\cdot s^{-2}}.

Explanation:

Let \omega denote the required angular velocity of this Ferris wheel. Let m denote the mass of a particular passenger on this Ferris wheel.

At the topmost point of the Ferris wheel, there would be at most two forces acting on this passenger:

  • Weight of the passenger (downwards), m\, g, and possibly
  • Normal force F_\text{normal} that the Ferris wheel exerts on this passenger (upwards.)

This passenger would feel "weightless" if the normal force on them is 0- that is, F_\text{normal} = 0.

The net force on this passenger is (m\, g - F_\text{normal}). Hence, when F_\text{normal} = 0, the net force on this passenger would be equal to m\, g.

Passengers on this Ferris wheel are in a centripetal motion of angular velocity \omega around a circle of radius r. Thus, the centripetal acceleration of these passengers would be a = \omega^{2}\, r. The net force on a passenger of mass m would be m\, a = m\, \omega^{2}\, r.

Notice that m\, \omega^{2} \, r = (\text{Net Force}) = m\, g. Solve this equation for \omega, the angular speed of this Ferris wheel. Since g = 9.81\; {\rm m\cdot s^{-2}} and r = 23\; {\rm m}:

\begin{aligned} \omega^{2} = \frac{g}{r}\end{aligned}.

\begin{aligned} \omega &= \sqrt{\frac{g}{r}} \\ &= \sqrt{\frac{9.81\; {\rm m \cdot s^{-2}}}{23\; {\rm m}}} \\ &\approx 0.653\; {\rm rad \cdot s^{-1}} \end{aligned}.

The question is asking for the angular velocity of this Ferris wheel in the unit {\rm rpm}, where 1\; {\rm rpm} = (2\, \pi\; {\rm rad}) / (60\; {\rm s}). Apply unit conversion:

\begin{aligned} \omega &\approx 0.653\; {\rm rad \cdot s^{-1}} \\ &= 0.653\; {\rm rad \cdot s^{-1}} \times \frac{1\; {\rm rpm}}{(2\, \pi\; {\rm rad}) / (60\; {\rm s})} \\ &= 0.653\; {\rm rad \cdot s^{-1} \times \frac{60\; {\rm s}}{2\, \pi\; {\rm rad}} \times 1\; {\rm rpm} \\ &\approx 6.2\; {\rm rpm} \end{aligned}.

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