Answer:
sorry- but what........?!
Answer:
A <em>concave</em><em> </em><em>lens</em><em> </em><em>is</em><em> </em><em>thinner</em><em> </em><em>at</em><em> </em><em>the</em><em> </em><em>cen</em><em>ter</em><em> </em><em>and </em><em>thick</em><em>er</em><em> </em><em>at</em><em> </em><em>the</em><em> </em><em>edges</em><em> </em><em>while</em><em> </em><em>a</em><em> </em><em>convex </em><em>lens </em><em>is</em><em> </em><em>thicker</em><em> </em><em>at</em><em> </em><em>the</em><em> </em><em>centre</em><em> </em><em>and</em><em> </em><em>thinner</em><em> </em><em>at</em><em> </em><em>the</em><em> edges</em><em>.</em>
(c) as the change in the dependent variable is in direct CORRELATION to the change in the independent variable.
Unbalanced forces is what they are called
T is in seconds (s)
<span>2pi is dimensionless </span>
<span>L is in meters (m) </span>
<span>g is in meters per second squared (m/s^2) </span>
<span>so you can write the equation for the period of the simple pendulum in its units... </span>
<span>s=sqrt(m/(m/s^2)) </span>
<span>simplify</span>
<span>s=sqrt(m*s^2*1/m) cancelling the m's </span>
<span>s=sqrt(s^2) </span>
<span>s=s </span>
<span>therefore the dimensions on the left side of the equation are equal to the dimensions on the right side of the equation.</span>
