Question 16 webassign.net Parad se Valley Homework 6: Chapt... Baby Boom AP U Ap

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Question 16

webassign.net Parad se Valley Homework 6: Chapt... Baby Boom AP U Ap study notes All Unread 15. 1/1 pointsI Previous Answers Mi4 6.8.049. My Notes Ask Your You throw a ball straight up, and it reaches a height of 19 m above your hand before falling back down. What was the speed of the ball just after it left your hand? 19.298 Pickleball Rules Summary albanypickleball.com Pickleball Rules Summary and ink to the Official Pickl... m/s Additional Materials Section 6.8 webassign.net LAB RESOURCES Linear Regression Percent Error a... The Common Application The Common Application Welcome, Chil First Year C. 16. 0/1 points Previous Answers Mi4 6.8.051 My Notes Ask Your An object with mass 3 kg moves from a location 22, 25,-41) m near the Earth's surface to location (-27, 11, 50) m, what is the change in the potential energy of the system consisting of the object plus the Earth? (Assume that the ty axis points vertically up from the ground.) Skip to main content Chi Nguyen Page path Dashbo... Enter a number Skip to main content Chi Section 6.8 Nguyen Page path Dashbo... chi nguyen23 | Quizlet Quizlet is a lightning fast Submit Answer Save Progress Practice Another Version way to learn vocabulary Distance and 17. 2/2 points | Previous Answers Mi4 6.8.053. My Notes Ask Your Physics Tutorial 1-D Kinematics Newton's Laws... Use energy conservation to find the approximate final speed of a basketball dropped from a height of 1.95 m (roughly the height of a professional basketball player) 6.182 m/s Why don't you need to know the mass of the basketball The mass is not important because the ball starts from rest. The kinetic and potential energies of the ball are each proportional to its mass, so the mass term cancels out. The mass of the basketball is small enough compared to that of the basketball player that it is negligible

Explanation / Answer

15) v^2 - u^2 = 2gH

0 - u^2 = 2*-9.8*19

u = 19.3 m/s

so the initial velocity was 19.3 m/s

16) chnage in height = 25 -11 = 14

Potential energy = mgH = -3*9.8*14 = -411.6 J

17) KE = Potential

0.5mv^2 = mgh

v = sqrt(2gh)

v = sqrt(2*9.8*1.95)

= 6.18 m/s

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