<p>The formula are
a=V^(2)/r
T=rFsinθ</p>
<ol>
<li>Given the follow such that v is 5 m/s and r is 2 meters, find a.</li>
<li>Originally, a=12 m/s^2, then r is doubled. Find the new value of a.</li>
<li>Use the second equation to find θ when T=4 Nm, r=2m, and F=10N.</li>
</ol>
<p>Thank You</p>
<p>Not there yet (is this Physics C or B?), though given the equations (the first being rotational acceleration and the second being torque), it’s not too bad.</p>
<ol>
<li><p>Plugin the values into the formula. a = 5^2/2 = 25/2 (12.5)</p></li>
<li><p>Consider the fact that a is inversely proportional to r. So if a ~ 1/r. If r is doubled, a is halved. Therefore, the new value of a is 6 m/s^2.</p></li>
<li><p>4 = 2<em>10sinθ. → 4/(2</em>10) = sinθ —> 4/20 = sin θ → 1/5 = sin θ.
θ = arcsin(1/5) = 11.54 degrees.</p></li>
</ol>