Surface area and mass. The surface area of a person whose mass is 75 kg can be approximated by f ( h ) = 0.144 h 1 / 2 , where f ( h ) is measured in square meters and h is the person's height in centimeters ( Source· US Oncology. ) a. Find the approximate surface area of a person whose mass is 75 kg and whose height is 180 cm. b. Find the approximate surface area of a person whose mass is 75 kg and whose height is 170 cm. c. Graph the function f for 0 ≤ h ≤ 200 .
Surface area and mass. The surface area of a person whose mass is 75 kg can be approximated by f ( h ) = 0.144 h 1 / 2 , where f ( h ) is measured in square meters and h is the person's height in centimeters ( Source· US Oncology. ) a. Find the approximate surface area of a person whose mass is 75 kg and whose height is 180 cm. b. Find the approximate surface area of a person whose mass is 75 kg and whose height is 170 cm. c. Graph the function f for 0 ≤ h ≤ 200 .
Solution Summary: The author explains the approximate surface area function, f(h)=0.144h raisebox1ex1!
The terminal velocity of a skydiver is reached typically about 10-12 seconds into a
dive.
The square of the terminal velocity of a skydiver (v) is directly proportional to the
skydiver's weight (w) and inversely proportional to the skydiver's ground-facing
surface area (A).
Write an equation that models this situation. Use k for the constant of proportionality.
Suppose Christy weighs 60 kilograms and has an approximate downward-facing area of 0.18 square meters.
Last week, Christy went skydiving and the terminal velocity was measured as 241 kilometers per hour.
If another person goes skydiving under similar conditons and they weigh 82 kilograms with a ground-facing
surface area of 0.33 square meters, what would be their terminal velocity?
kilometers per hour {Give at least 3 decimal place accuracy}
A trough has ends shaped like isosceles triangles, with width of w = 4 meters and height h = 6 meters, and the trough is L = 13 meters long. Water is
being pumped into the trough at a rate of 7 cubic meters per minute. At what rate does the height of the water change when the water is 1 meters deep?
W
1.615 m/min. help (numbers)
For a moving object, velocity V, distance S, and acceleration A, are related by the formula below. Find a when V = 30m/sec and S = 50m. V^2 = 2AS
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, calculus and related others by exploring similar questions and additional content below.
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