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January 27, 2015

January 27, 2015

Posted by **mike** on Monday, February 15, 2010 at 7:26pm.

KE = (.5)(m)(v^2)

209.3 J = (.5)(.25kg)(v^2)

1674.4 = v^2 **square root both sides**

v = 40.92

^Is ANY of that correct? If not, can anyone help? Thanks so much

- Physics -
**drwls**, Monday, February 15, 2010 at 8:54pmI don't see what 1/4 kg of water has to do with a baseball, but you did the problem correctly.

The specific heat of a baseball is different from that of water, and its mass is 1/7 kg, so what you have calculated in NOT the velocity needed to raise the baseball 0.2 degrees. It is an arbitrary number.

- Physics -
**Tiffany**, Tuesday, January 17, 2012 at 5:30pmyes that is correct

- Physics -
**Jaeda**, Monday, February 27, 2012 at 9:15pmYou are right!!

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