Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
expand_more
expand_more
format_list_bulleted
Concept explainers
Videos
Question
Chapter 5, Problem 133P
To determine
To Estimate: Terminal speed of flying squirrel.
Whether calculations support Sally’s claim or not.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
Callie is running a 400 m race around a 400 m track. On the backstretch her velocity is 8 m/s, but she is running into a 2 m/s headwind. How large is the drag force that acts on Callie? Assume that the density of the air is 1.2 kg/m3, that Callie’s cross-sectional area is 0.5 m2, and that her coefficient of drag is 1.1.
A car has a drag coefficient Cd
kg.m-3.
= 0.30, a frontal area of A= 1.9 m² and a mass 1.2 tonnes. The density of air is 1.2
i)
What is the drag force when it is traveling at v = 110 kph in a straight line?
Fdrag
N.
Hint: Retain accurate values until the final calculation, but remember significant figures.
Enter answer here
A 565-g squirrel with a surface area of 860 cm2 falls from a 6.0-m tree to the ground. Estimate its terminal velocity. (Use the drag coefficient for a horizontal skydiver. Assume that the cross-sectional area of the squirrel can be approximated as a rectangle of width 11.1 cm and length 22.2 cm. Note, the squirrel may not reach terminal velocity by the time it hits the ground. Give the squirrel's terminal velocity, not it's velocity as it hits the ground.)
-------m/s
What will be the velocity of a 56.5-kg person hitting the ground, assuming no drag contribution in such a short distance?
-------m/s
Chapter 5 Solutions
Physics for Scientists and Engineers
Ch. 5 - Prob. 1PCh. 5 - Prob. 2PCh. 5 - Prob. 3PCh. 5 - Prob. 4PCh. 5 - Prob. 5PCh. 5 - Prob. 6PCh. 5 - Prob. 7PCh. 5 - Prob. 8PCh. 5 - Prob. 9PCh. 5 - Prob. 10P
Ch. 5 - Prob. 11PCh. 5 - Prob. 12PCh. 5 - Prob. 13PCh. 5 - Prob. 14PCh. 5 - Prob. 15PCh. 5 - Prob. 16PCh. 5 - Prob. 17PCh. 5 - Prob. 18PCh. 5 - Prob. 19PCh. 5 - Prob. 20PCh. 5 - Prob. 21PCh. 5 - Prob. 22PCh. 5 - Prob. 23PCh. 5 - Prob. 24PCh. 5 - Prob. 25PCh. 5 - Prob. 26PCh. 5 - Prob. 27PCh. 5 - Prob. 28PCh. 5 - Prob. 29PCh. 5 - Prob. 30PCh. 5 - Prob. 31PCh. 5 - Prob. 32PCh. 5 - Prob. 33PCh. 5 - Prob. 34PCh. 5 - Prob. 35PCh. 5 - Prob. 36PCh. 5 - Prob. 37PCh. 5 - Prob. 38PCh. 5 - Prob. 39PCh. 5 - Prob. 40PCh. 5 - Prob. 41PCh. 5 - Prob. 42PCh. 5 - Prob. 43PCh. 5 - Prob. 44PCh. 5 - Prob. 45PCh. 5 - Prob. 46PCh. 5 - Prob. 47PCh. 5 - Prob. 48PCh. 5 - Prob. 49PCh. 5 - Prob. 50PCh. 5 - Prob. 51PCh. 5 - Prob. 52PCh. 5 - Prob. 53PCh. 5 - Prob. 54PCh. 5 - Prob. 55PCh. 5 - Prob. 56PCh. 5 - Prob. 57PCh. 5 - Prob. 58PCh. 5 - Prob. 59PCh. 5 - Prob. 60PCh. 5 - Prob. 61PCh. 5 - Prob. 62PCh. 5 - Prob. 63PCh. 5 - Prob. 65PCh. 5 - Prob. 67PCh. 5 - Prob. 68PCh. 5 - Prob. 69PCh. 5 - Prob. 70PCh. 5 - Prob. 71PCh. 5 - Prob. 72PCh. 5 - Prob. 73PCh. 5 - Prob. 74PCh. 5 - Prob. 75PCh. 5 - Prob. 76PCh. 5 - Prob. 77PCh. 5 - Prob. 78PCh. 5 - Prob. 79PCh. 5 - Prob. 80PCh. 5 - Prob. 82PCh. 5 - Prob. 83PCh. 5 - Prob. 84PCh. 5 - Prob. 85PCh. 5 - Prob. 86PCh. 5 - Prob. 87PCh. 5 - Prob. 88PCh. 5 - Prob. 89PCh. 5 - Prob. 90PCh. 5 - Prob. 91PCh. 5 - Prob. 92PCh. 5 - Prob. 93PCh. 5 - Prob. 94PCh. 5 - Prob. 95PCh. 5 - Prob. 96PCh. 5 - Prob. 97PCh. 5 - Prob. 101PCh. 5 - Prob. 102PCh. 5 - Prob. 103PCh. 5 - Prob. 104PCh. 5 - Prob. 105PCh. 5 - Prob. 106PCh. 5 - Prob. 107PCh. 5 - Prob. 108PCh. 5 - Prob. 109PCh. 5 - Prob. 110PCh. 5 - Prob. 111PCh. 5 - Prob. 112PCh. 5 - Prob. 113PCh. 5 - Prob. 114PCh. 5 - Prob. 115PCh. 5 - Prob. 116PCh. 5 - Prob. 117PCh. 5 - Prob. 118PCh. 5 - Prob. 119PCh. 5 - Prob. 120PCh. 5 - Prob. 121PCh. 5 - Prob. 122PCh. 5 - Prob. 123PCh. 5 - Prob. 124PCh. 5 - Prob. 125PCh. 5 - Prob. 126PCh. 5 - Prob. 127PCh. 5 - Prob. 128PCh. 5 - Prob. 129PCh. 5 - Prob. 130PCh. 5 - Prob. 131PCh. 5 - Prob. 132PCh. 5 - Prob. 133PCh. 5 - Prob. 134PCh. 5 - Prob. 135PCh. 5 - Prob. 136PCh. 5 - Prob. 137PCh. 5 - Prob. 138PCh. 5 - Prob. 139P
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.Similar questions
- The drag coefficient C in FD=12CAv2 (Eq. 6.5) depends primarily on the shape of the object. You already have developed an intuition about what shapes correspond to a low C by observing the shapes of aerodynamic cars, boats, and even bullets. Which object, a sphere or a cube, would have a larger drag coefficient, assuming they are nearly the same size? Explain your reasoning. What aspect of an object most determines its drag coefficient?arrow_forwardA car is a height of 1.6 meters and width of 2 meters. (the length is immaterial). The following is an actual excerpt from the manufacturers specifications: Fuel capacity: 51 liters/ 13.5 US gal/ 11.2 gal Drag coefficient 0.42 (at highway speed) a) If I am driving at 100 km/hr through air what is the drag force on the car. Note this is the same as if the car were stationary in at 100 km/hr wind. Air density is 1.2 kg/m^3 viscosity is 0.018 cparrow_forwardA guy dives from a cliff of height H at an initial speed 0 (for simplicity). During the first phase of his fall, he wraps his arms. During the second phase, he stretches his arms to slow himself down so that his touchdown speed at the water surface is vo. The air drag coefficient during the first phase is k while it is ßk during the second phase. The air drag force is always proportional to the falling speed. At what height the guy should open his arms so that his touchdown speed is vo? Compute the total falling time.arrow_forward
- A raindrop with a radius R= 1.2 mm falls from a cloud that is at height h = 1200 m above the ground. The drag coefficient C for the drop is 0.40. Assume that the drop is spherical throughout its fall. The density of water ρw is 1000 kg/m3, and the density of air ρa is 1.1 kg/m3. If you know that the raindrop reaches terminal speed after falling just a few meters. What is the terminal speed?arrow_forwardMany birds can attain very high speeds when diving. Using radar, scientists measured the altitude of a barn swallow in a vertical dive; it dropped 208 m in 3.0 s. The mass of the swallow was estimated to be 0.018 kg, and its cross-section area as 5.6 x 10-4 m2. What was the drag coefficient for this swallow as it dove?arrow_forward"A raindrop with radius R= 1.6 mm falls from a cloud that is at height h = 1200 m above the ground. The drag coefficient C for the drop is 0.60. Assume that the drop is spherical throughout its fall. The density of water ρw is 1000 kg/m3, and the density of air ρa is 1.2 kg/m3. If you know that the raindrop reaches terminal speed after falling just a few meters. What is the terminal speed?" a 7.6 m/s b 9.8 m/s c 4.6 m/s d The data is not enough to calculate the speedarrow_forward
- The drag on a pitched baseball can be surprisingly large. Suppose a 145 gram baseball with a diameter of 7.4 cm has an initial speed of 40.2 m/s (90 mph). a) What is the magnitude of the ball’s acceleration due to the drag force? b) If the ball had this same acceleration during its entire 18.4m trajectory, what would its final speed be?arrow_forwardCalculate the ratio of the drag force on a jet flying at 900 km/h at an altitude of 10 km to the drag force on a prop-driven transport flying at one-fifth the speed and half the altitude of the jet. The density of air is 0.38 kg/m3 at 10 km and 0.67 kg/m3 at 5.0 km. Assume that the airplanes have the same effective cross-sectional area and drag coefficient C. (drag on jet/drag on transport) =arrow_forwardThe maximum ('terminal') velocity, vt, that a falling object achieves depends on its mass m, the acceleration g due to gravity, the density p of the air, the object's frontal area A, and the object's drag coefficient C, which is a dimensionless quantity. The equation for the terminal velocity is: 2mg V = pCA Part (a): A skydiver jumps out of an airplane (regrettably) without his parachute. To slow his descent, he orients himself in a 'spread eagle' position, thereby increasing his area A to 0.70 m?. His drag coefficient Cis 1.0, which is a unitless quantity. Calculate his terminal velocity in m/s if his mass is 85 kg, the air density p is 1.21 kg/m3, and the constant g is 9.80 m/s?. Show your work, and ensure that each numeric value has the appropriate units. Your answer should be rounded to the appropriate number of significant digits. Part (b): Convert the answer from part (a) to miles per hour using the dimensional analysis method we reviewed in class. Show your work, and ensure…arrow_forward
- If P = 200 N, determine the friction developed between the 50-kg crate and the ground. The coefficient of static friction between the crate and the ground is us= 0.3.arrow_forwardConsider the heaviest box that you can push at constant speed across a level floor, where the coefficient of kinetic friction is 0.50, and estimate the maximum horizontal force that you can apply to the box. A box sits on a ramp that is inclined at an angle of 60 above the horizontal. The coefficient of kinetic friction between the box and the ramp is 0.50. If you apply the same magnitude force, now parallel to the ramp, that you applied to the box on the floor, what is the heaviest box (in pounds) that you can push up the ramp at constant speed? (In both cases assume you can give enough extra push to get the box started moving.)arrow_forwardA tennis ball is thrown at 25 m/s on a day when the wind speed is 0 m/s. What is the drag force acting on the tennis ball if the coefficient of drag is 0.4 and the reference area is 0.005 m2? The density of air is 1.2 kg/m3.arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Physics for Scientists and Engineers: Foundations...PhysicsISBN:9781133939146Author:Katz, Debora M.Publisher:Cengage Learning
Physics for Scientists and Engineers: Foundations...
Physics
ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Cengage Learning
Newton's Third Law of Motion: Action and Reaction; Author: Professor Dave explains;https://www.youtube.com/watch?v=y61_VPKH2B4;License: Standard YouTube License, CC-BY