EBK BASIC BIOMECHANICS
7th Edition
ISBN: 9780100409453
Author: Hall
Publisher: YUZU
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Chapter 10, Problem 4AP
A buoy marking the turn in the ocean swim leg of a triathlon becomes unanchored. If the current carries the buoy southward at 0.5 m/s, and the wind blows the buoy westward at 0.7 m/s, what is the resultant displacement of the buoy after 5 min? (Answer: 258m; ∠ = 54.5° west of due south)
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Chapter 10 Solutions
EBK BASIC BIOMECHANICS
Ch. 10 - A runner completes 6 laps around a 400 m track...Ch. 10 - A ball rolls with an acceleration of 0.5 m/s2. If...Ch. 10 - A wheelchair marathoner has a speed of 5 m/s after...Ch. 10 - An orienteer runs 400 m directly east and then 500...Ch. 10 - An orienteer runs north at 5 m/s for 120 s and...Ch. 10 - Why are the horizontal and vertical components of...Ch. 10 - A soccer ball is kicked with an initial horizontal...Ch. 10 - If a baseball, a basketball, and a 71.2-N shot...Ch. 10 - A tennis ball leaves a racket during the execution...Ch. 10 - Prob. 10IP
Ch. 10 - Answer the following questions pertaining to the...Ch. 10 - Provide a trigonometric solution for Introductory...Ch. 10 - Provide a trigonometric solution for Introductory...Ch. 10 - A buoy marking the turn in the ocean swim leg of a...Ch. 10 - A sailboat is being propelled westerly by the wind...Ch. 10 - A Dallas Cowboy carrying the ball straight down...Ch. 10 - A soccer ball is kicked from the playing field at...Ch. 10 - A ball is kicked a horizontal distance of 45.8 m....Ch. 10 - A badminton shuttlecock is struck by a racket at a...Ch. 10 - An archery arrow is shot with a speed of 45 m/s at...
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- What is the relationship between Net force, mass and acceleration. Newton’s second law of motion?arrow_forwardUse the following information to answer questions 20 and 21. Assignment Booklet 4B Two cars, each with a mass of 1000 kg, are travelling in opposíte directionsn is car travelling to the right is travelling 30 m/s, and the car travelling to the lert is travelling 20 m/s. 1000 kg 30 m/s 1000 kg 20 m/s 20. What is the total momentum of the vehicles after they collicde? A. -50 000 kg-m/s B. 50 000 kg.m/s C. -10 000 kg.m/s D. 10 000 kg.m/s al ne 21. If the two vehicles collide and lock together, what is their velocity after the collision? A. -5 m/s Aon s quAua B. 5 m/s elg C. -10 m/s D. 10 m/s Return to page 70 of the Student Module Booklet and begin the Section 3 Review. os elomun ef et birov ort to solniedarrow_forwardWhich of the following components of Km is largest in magnitude? k2 O k-1 O k1 O k-2arrow_forward
- please provide an example to solve for Velocity using the Michaelis-Menten Equation ?arrow_forwarda toy car rolls 10 meters (m) across the floor. it takes 5 seconds (s) to cross this distance. what is the speed of this car?arrow_forwardBased on the speculations of Nicole Oresme, and on the equation relating spatial distance, time, initial velocity, and constant acceleration developed by the Mertonian Calculators, how far would you expect a falling object to travel in 5 seconds (falling from a spaceship towards the Earth, in the vacuum of space), starting at 0 feet/second, with constant acceleration (32 feet/sec2), and neglecting possible air friction? 144 feet 256 feet 400 feet 576 feet 784 feetarrow_forward
- A scientist was investigating if differences in the frictional work performed on a model car can change depending on its mass (in grams) and whether the car moves up or down an inclined plane. They decided to measure the amount of frictional force experienced by the model car and the distance it traveled in meters. The scientists were able to evaluate the frictional work using the following data. Mass (g) Distance (m) Force Work Done by Friction (J) car going up the incline 100 39 0.063 2.457 car going down the incline 70 39 0.2309 ? It is known that the relationship between force and distance determines the work done by friction (W+). W₁ = fd Wf work done by friction f = force d = distance Question: How much work done by friction was exerted on the car as it moved down the inclined plane? You may use a calculator. 1 2.457 9.005 11.46 16.16 PREVIOUS FINISHarrow_forwardKepler’s third law can be stated mathematically as: velocity = distance/time (years)2 = (A.U.)3 S = vot + ½ (a) t2 vm = ½ (vo + vf) F = GM1M2/d2arrow_forwardThe “mean-speed theorem” for finding average velocity under constant acceleration, proposed by the Oxford Calculators, and demonstrated geometrically by Nicole Oresme, is expressed algebraically as: density = weight/volume (m1)(v1) = (m2)(v2) (vm) = 1/2 (v0 + vf) s = (v0)(t) + 1/2 (a)(t2) velocity = distance/timearrow_forward
- How would warmer temperatures affect Km and Vmax on Michaelis-Menton curve?arrow_forwardRotate the ball horizontally on an 80 cm long non-stretchable cord with angular velocity of 3 s^(-1). After ten seconds of clockwise rotation, the cord breaks. At what speed and in what direction does the ball fly, if it was faced north at time t = Os? Where and after how much time does the ball land on the ground that is 1 m below the plaine in which we rotate the string? {Solution: v= (0.37 m/s,2.37 m/s); d=1.07 m, t=0.447 s.) }arrow_forwardThe “mean-speed theorem” for calculating average velocity under constant acceleration, developed by Thomas Bradwardine and the Mertonian Calculators at Oxford University, is expressed algebraically as: density = weight/volume (m1)(v1) = (m2)(v2) C. (vm) = 1/2 (v0 + vf) s = (v0)(t) + 1/2 (a)(t2) velocity = distance/timearrow_forward
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