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antoniya [11.8K]
3 years ago
13

A 20 N force is applied to a .170kg hockey puck. What is the acceleration of the puck?

Physics
1 answer:
shutvik [7]3 years ago
4 0
You want to use the equation F=ma, so
20N=(0.170kg)a, divide 0.170 from both sides
a=117.64705..., since .170 has 3 significant figures, round to 3 significant figures
a=118 m/s²





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pickupchik [31]
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<span>In that particular situation, you can prove it like this: </span>

<span>initial velocity is Vo </span>
<span>launch angle is α </span>

<span>initial vertical velocity is </span>
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<span>t = time = ? </span>
<span>a = acceleration by gravity = g (= -9.8 m/s²) </span>
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<span>0 = Vo×sin(α)×t + g×t²/2 </span>
<span>0 = (Vo×sin(α) + g×t/2)×t </span>
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<span>t = -2×Vo×sin(α)/g </span>

<span>Now look at the horizontal range. </span>
<span>r = v × t </span>
<span>where </span>
<span>r = horizontal range = ? </span>
<span>v = horizontal velocity = Vh = Vo×cos(α) </span>
<span>t = time = -2×Vo×sin(α)/g </span>
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<span>r = (Vo×cos(α)) × (-2×Vo×sin(α)/g) </span>
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<span>To find the extreme values of r (minimum or maximum) with variable α, you must find the first derivative of r with respect to α, and set it equal to 0. </span>

<span>dr/dα = d[-(Vo)²×sin(2α)/g] / dα </span>
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<span>cos(2α) = 0 </span>
<span>2α = 90° </span>
<span>α = 45° </span>
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