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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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A bridge is supported by two piers located 14 meters apart. Both the left and right piers provide an upward force on the bridge,
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Answer:

4375 N, 7875 N

Explanation:

since the body is equilibrium

total upward force = total downward force

w weight of the bridge = FL + FR

when the car was introduced,

total downward force = total upward force

FL₁ + FR₁ = w + (m × acceleration due to gravity) with w = FL + FR

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taken the left as the pivot point and using the principle of moment

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ΔFL = 12250 - 4375 = 7875 N

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Answer:

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