The PE of the box that is on a 2.0 m high self is 1600 J. What is the power expelled to lift the box to this height in 10.0 seco
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The food package will strike the ground at 11 degrees below the horizontal.
<h3>Time for the food package to hit the ground</h3>The time for the food package to hit the ground is calculated as follows;
h = vt + ¹/₂gt²
<em>let the initial velocity be horizontal</em>
4900 = 0(t) + (0.5 x 9.8)t²
4900 = 4.9t²
t² = 4900/4.9
t² = 1,000
t = √1,000
t = 31.62 s
<h3> Final speed of the food package when it hits ground</h3>vf(y) = vo + gt
vf(y) = 0 + (31.62 x 9.8)
vf(y) = 309.88 m/s
<h3>Angle of projection</h3>The horizontal component of the speed will be constant, while vertical component will change
Angle below the horizontal = 90 - 79 = 11⁰
Thus, the food package will strike the ground at 11 degrees below the horizontal.
Learn more about angle of projection here: brainly.com/question/10671136