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goldenfox [79]
3 years ago
9

A thin, uniform rod is hinged at its midpoint. to begin with, one-half of the rod is bent upward and is perpendicular to the oth

er half. this bent object is rotating at an angular velocity of 6.9 rad/s about an axis that is perpendicular to the left end of the rod and parallel to the rod's upward half (see the drawing). without the aid of external torques, the rod suddenly assumes its straight shape. what is the angular velocity of the straight rod?
Physics
1 answer:
bazaltina [42]3 years ago
5 0
<span>Answer: initial I = (m/2)L²/3 + (m/2)L² where L = ½ the length of the rod, and the vertical half can be treated as a point mass. initial I = mL²(1/6 + 1/2) = 2mL²/3 final I = m(2L)²/3 = 4mL²/3 Since I has doubled and momentum is conserved, ω has halved. ω = 3.9 rad/s. Formulaically: 2mL²/3 * 7.8rad/s = 4mL²/3 * ω</span>
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Explanation:

Formula which holds true for a leans with radii R_{1} and R_{2} and index refraction n is given as follows.

          \frac{1}{f} = (n - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}]

Since, the lens is immersed in liquid with index of refraction n_{1}. Therefore, focal length obeys the following.  

            \frac{1}{f_{1}} = \frac{n - n_{1}}{n_{1}} [\frac{1}{R_{1}} - \frac{1}{R_{2}}]  

             \frac{1}{f(n - 1)} = [\frac{1}{R_{1}} - \frac{1}{R_{2}}]

and,       \frac{n_{1}}{f(n - n_{1})} = \frac{1}{R_{1}} - \frac{1}{R_{2}}

or,          f_{1} = \frac{fn_{1}(n - 1)}{(n - n_{1})}

              f_{w} = \frac{10 \times 1.33 \times (1.56 - 1)}{(1.56 - 1.33)}

                          = 32.4 cm

Using thin lens equation, we will find the focal length as follows.

             \frac{1}{f} = \frac{1}{s_{o}} + \frac{1}{s_{i}}

Hence, image distance can be calculated as follows.

       \frac{1}{s_{i}} = \frac{1}{f} - \frac{1}{s_{o}} = \frac{s_{o} - f}{fs_{o}}

              s_{i} = \frac{fs_{o}}{s_{o} - f}

             s_{i} = \frac{32.4 \times 100}{100 - 32.4}

                       = 47.9 cm

Therefore, we can conclude that the focal length of the lens in water is 47.9 cm.

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It's a bit of a trick question, had the same one on my homework. You're given an electric field strength (1*10^5 N/C for mine), a drag force (7.25*10^-11 N) and the critical info is that it's moving with constant velocity(the particle is in equilibrium/not accelerating). 
<span>All you need is F=(K*Q1*Q2)/r^2 </span>
<span>Just set F=the drag force and the electric field strength is (K*Q2)/r^2, plugging those values in gives you </span>
<span>(7.25*10^-11 N) = (1*10^5 N/C)*Q1 ---> Q1 = 7.25*10^-16 C </span>
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