Question #34b47

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Answer 1 · 2016-02-04T11:16:48

$E_{k} = \frac{603, 2074011}{4, 18} = 144, 307990694 (Calories)$ $E_k=(603,2074011)/(4,18)=144,307990694 "(Calories)"$

$m = 3, 15 \cdot 10^{3} l b$ $m=3,15*10^3 lb$

$1 l b = 0, 45359237 k g$ $1 lb=0,45359237 kg$

$m = 3150 \cdot 0, 45359237 = 1428, 8259655 k g (convert lb to kg)$ $m= 3150*0,45359237=1428,8259655 kg"(convert lb to kg)"$

$1 mile=1,60934 km$ $1" mile=1,60934 km"$

$v=65 miles per hour$ $"v=65 miles per hour"$

$v = 65 \cdot 1, 60934 = 104, 6071 (km per hour)$ $v=65*1,60934=104,6071 "(km per hour)"$

$v=(104,6071*10cancel(00))/(36cancel(00))=29,0575277778 "(meter per second)"$

$E_k=1/2*m*v^2$

$E_k=((1428,8259655*29,0575277778^2)/2)$

$E_k=603,2074011 "Joule"$

$"1 Calorie=4,18 Joule"$

$E_k=(603,2074011)/(4,18)=144,307990694 "(Calories)"$