Which of the following is a helpful process in evaluating scientific claims?

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inna [3103]

Answer:

a = 18.28 ft/s²

Explanation:

the values provided are:

duration of force application, t= 10 s

Work done = 10 Btu

mass of the object = 15 lb

acceleration, a =? ft/s²

1 Btu = 778.15 ft.lbf

thus, 10 Btu = 7781.5 ft.lbf

$m = \dfrac{15}{32.174}\ slug$

m = 0.466 slug

So,

the work is equivalent to the change in kinetic energy

$W = \dfrac{1}{2} m (v_f^2-v_i^2)$

$7781.5 = \dfrac{1}{2}\times 0.466\times v_f^2$

$v_f = 182.75\ ft/s$

The acceleration of the object is therefore

$a = \dfrac{v_f-v_o}{t}$

$a = \dfrac{182.75-0}{10}$

a = 18.28 ft/s²

the constant acceleration of the object is calculated to be 18.28 ft/s²

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3 months ago