Use the given information to answer the questions. The force exerted by the wind on a plane surface varies jointly with the square of the velocity of the wind and with the area of the plane surface. If the area of the surface is 40 square feet surface and the wind velocity is 20 miles per hour, the resulting force is 15 pounds. Find the force on a surface of 65 square feet with a velocity of 30 miles per hour.
Added by Pedro P.
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The equation is: \[ f = Kw^2a \] where: - \( f \) is the force exerted by the wind, - \( K \) is the constant of proportionality, - \( w \) is the velocity of the wind, - \( a \) is the area of the plane surface. Show more…
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For the following exercises, use the given information to answer the questions. The force exerted by the wind on a plane surface varies jointly with the square of the velocity of the wind and with the area of the plane surface. If the area of the surface is 40 square feet surface and the wind velocity is 20 miles per hour, the resulting force is 15 pounds. Find the force on a surface of 65 square feet with a velocity of 30 miles per hour.
Polynomial and Rational Functions
Modeling Using Variation
The force exerted by the wind on a plane surface varies jointly with the square of the velocity of the wind and with the area of the plane surface. If the area of the surface is 20 square feet surface and the wind velocity is 30 miles per hour, the resulting force is 10 pounds. Find the force, F, on a surface of 75 square feet with a velocity of 40 miles per hour. (Round your answer to two decimal places.)
Nishant K.
During a Chicago storm, winds can whip horizontally at speeds of 120 $\mathrm{km} / \mathrm{h}$ . If the air strikes a person at the rate of 45 $\mathrm{kg} / \mathrm{s}$ per square meter and is brought to rest, calculate the force of the wind on a person. Assume the person is 1.60 $\mathrm{m}$ high and 0.50 $\mathrm{m}$ wide. Compare to the typical maximum force of friction $(\mu \approx 1.0)$ between the person and the ground, if the person has a mass of 75 $\mathrm{kg}$ .
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