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Part A = The integral with respect to time of a force applied to an object is a measure called impulse, and the impulse applied to an object during a time interval determines its change in momentum during the time interval. The safety of a t-shirt launcher, used to help get crowds cheering at baseball games, is being evaluated. As a first step in the evaluation, engineers consider the design momentum of the launched t-shirts. The springs in the launcher are designed to apply a variable force to a t-shirt over a time interval of t 0.5 s. The force as a function of time is given by F(t) = at² + b, where a = −28 N/s² and b = 7.0 N. The momentum of the t-shirt will be its initial momentum (po = 0) plus its change in momentum due to the applied impulse: pf = po+ St-F(t) dt. By applying the given time dependent function for F(t) and performing the integration, which of the following is the correct expression for pf? ▸ View Available Hint(s) rt=tL 0+(1/3+ ³ + b) 10/20 tL Pf= ○ tL 0+ (13) 기함 a 0+(3+bt) 0+(을)기향 Submit Previous Answers Part B Correct We check that we have obtained the correct form of the integral by performing differentiation of 1/3 +3 + bt with respect to t, which gives at² + b = F(t) as expected. The units of the momentum of the t-shirt are the units of the integral Stat F(t) dt, where F(t) has units of N and t has units of s. Given that 1 N = 1 kg • m/s², the units of momentum are: ▸ View Available Hint(s) kg. m/s³ kg. m/s kg. m/s² kg/s

Part A = The integral with respect to time of a force applied to an object is a measure called impulse, and the impulse applied to an object during a time interval determines its change in momentum during the time interval. The safety of a t-shirt launcher, used to help get crowds cheering at baseball games, is being evaluated. As a first step in the evaluation, engineers consider the design momentum of the launched t-shirts. The springs in the launcher are designed to apply a variable force to a t-shirt over a time interval of t 0.5 s. The force as a function of time is given by F(t) = at² + b, where a = −28 N/s² and b = 7.0 N. The momentum of the t-shirt will be its initial momentum (po = 0) plus its change in momentum due to the applied impulse: pf = po+ St-F(t) dt. By applying the given time dependent function for F(t) and performing the integration, which of the following is the correct expression for pf? ▸ View Available Hint(s) rt=tL 0+(1/3+ ³ + b) 10/20 tL Pf= ○ tL 0+ (13) 기함 a 0+(3+bt) 0+(을)기향 Submit Previous Answers Part B Correct We check that we have obtained the correct form of the integral by performing differentiation of 1/3 +3 + bt with respect to t, which gives at² + b = F(t) as expected. The units of the momentum of the t-shirt are the units of the integral Stat F(t) dt, where F(t) has units of N and t has units of s. Given that 1 N = 1 kg • m/s², the units of momentum are: ▸ View Available Hint(s) kg. m/s³ kg. m/s kg. m/s² kg/s

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Part A = The integral with respect to time of a force applied to an object is a measure called impulse, and the impulse applied to an object during a time interval determines its change in momentum during the time interval. The safety of a t-shirt launcher, used to help get crowds cheering at baseball games, is being evaluated. As a first step in the evaluation, engineers consider the design momentum of the launched t-shirts. The springs in the launcher are designed to apply a variable force to a t-shirt over a time interval of t 0.5 s. The force as a function of time is given by F(t) = at² + b, where a = −28 N/s² and b = 7.0 N. The momentum of the t-shirt will be its initial momentum (po = 0) plus its change in momentum due to the applied impulse: pf = po+ St-F(t) dt. By applying the given time dependent function for F(t) and performing the integration, which of the following is the correct expression for pf? ▸ View Available Hint(s) rt=tL 0+(1/3+ ³ + b) 10/20 tL Pf= ○ tL 0+ (13) 기함 a 0+(3+bt) 0+(을)기향 Submit Previous Answers Part B Correct We check that we have obtained the correct form of the integral by performing differentiation of 1/3 +3 + bt with respect to t, which gives at² + b = F(t) as expected. The units of the momentum of the t-shirt are the units of the integral Stat F(t) dt, where F(t) has units of N and t has units of s. Given that 1 N = 1 kg • m/s², the units of momentum are: ▸ View Available Hint(s) kg. m/s³ kg. m/s kg. m/s² kg/s
Transcribed Image Text:Part A = The integral with respect to time of a force applied to an object is a measure called impulse, and the impulse applied to an object during a time interval determines its change in momentum during the time interval. The safety of a t-shirt launcher, used to help get crowds cheering at baseball games, is being evaluated. As a first step in the evaluation, engineers consider the design momentum of the launched t-shirts. The springs in the launcher are designed to apply a variable force to a t-shirt over a time interval of t 0.5 s. The force as a function of time is given by F(t) = at² + b, where a = −28 N/s² and b = 7.0 N. The momentum of the t-shirt will be its initial momentum (po = 0) plus its change in momentum due to the applied impulse: pf = po+ St-F(t) dt. By applying the given time dependent function for F(t) and performing the integration, which of the following is the correct expression for pf? ▸ View Available Hint(s) rt=tL 0+(1/3+ ³ + b) 10/20 tL Pf= ○ tL 0+ (13) 기함 a 0+(3+bt) 0+(을)기향 Submit Previous Answers Part B Correct We check that we have obtained the correct form of the integral by performing differentiation of 1/3 +3 + bt with respect to t, which gives at² + b = F(t) as expected. The units of the momentum of the t-shirt are the units of the integral Stat F(t) dt, where F(t) has units of N and t has units of s. Given that 1 N = 1 kg • m/s², the units of momentum are: ▸ View Available Hint(s) kg. m/s³ kg. m/s kg. m/s² kg/s
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