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September 16, 2014

September 16, 2014

Posted by **Chad** on Wednesday, August 8, 2012 at 2:11pm.

of her breath as a function of the distance along the length of the straw can be modeled as 10-50x^2-70x^3

where force is in newtons and x is in meters. (a) How much work is done by her breath on the spitball as it travels the length of the barrel? (b) Assuming negligible friction and the straw is held horizontally, what is the speed of the spitball as it leaves the straw?

- Physics -
**Henry**, Wednesday, August 8, 2012 at 10:51pmF = 10 - 50x^2 - 70x^3.

F=10 - 50*(0.25)^2 - 70*(0.25)^3=5.78 N.

a. W = F * d = 5.78 * 0.25=1.45 Joules.

b. a = F/m = 5.78 / 0.012 = 481.7 m/s^2.

V^2 = Vo + 2a*d.

V^2 = 0 + 963.4*0.25 = 240.85

V = 15.5 m/s.

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