I believe that B is the answer.
Because the gravitational constant is not same as on earth
Answer:
Energy transfer is the movement of energy from one location to another. Energy transformation is when energy changes from one type to another. While energy can be transferred or transformed, the total energy always remains the same.
Explanation:
Cathy Rigby had a eating disorder.
The moment the stick comes to rest at θ=62.1° from horizontal.
<span>Angular acceleration = (net torque) / (moment of inertia) </span>
<span>α = τ/I </span>
<span>We have to add up the torques due to the bugs and the stick; and add up the moments of inertia due to all three also. </span>
<span>Let L be the stick's length and let m be the stick's mass (so "2.75m" is each bug's mass). And let's say the "lower" ladybug is on the left. Then the lower ladybug exerts this much torque: </span>
<span>τ_lowerbug = −(2/5)L(2.75mg)cosθ (negative because I am (arbitrarily) choosing counter-clockwise as the negative angular direction). </span>
<span>The upper ladybug exerts this much torque: </span>
<span>τ_upperbug = +(3/5)L(2.75mg)cosθ </span>
<span>The weight of the stick can be assumed to act through its center, which is 1/10 of the way from the fulcrum. So the stick exerts this much torque: </span>
<span>τ_stick = +(1/10)L(mg)cosθ </span>
<span>The net torque is thus: </span>
<span>τ_net = τ_lowerbug + τ_upperbug + τ_stick </span>
<span>= −(2/5)L(2.75mg)cosθ + (3/5)L(2.75mg)cosθ + (1/10)L(mg)cosθ </span>
<span>= (2.75(3/5−2/5)+1/10)(mgL)cosθ </span>
<span>Now for the moments of inertia. The bugs can be considered point masses of "2.75m" each. So for each of them you can use the simple formula: I=mass×R²: </span>
<span>I_lowerbug = (2.75m)((2/5)L)² = (2.75m)(4/25)L² </span>
<span>I_upperbug = (2.75m)((3/5)L)² = (2.75m)(9/25)L² </span>
<span>For the stick, we can use the parallel axis theorem. This says, when rotating something about an axis offset a distance "R" from its center of mass, the moment of inertia is: </span>
<span>I = I_cm + mR² </span>
<span>We know that for a stick about its center of mass, I_cm is (1/12)mL² (see many sources). And in this problem we know that it's offset by R=(1/10)L. So: </span>
<span>I_stick = (1/12)mL² + m((1/10)L)² </span>
<span>= (1/12)mL² + (1/100)mL² </span>
<span>= (7/75)mL² </span>
<span>So the total moment of inertia is: </span>
<span>I_total = I_lowerbug + I_upperbug + I_stick </span>
<span>= (2.75m)(4/25)L² + (2.75m)(9/25)L² + (7/75)mL² </span>
<span>= (2.75(4/25+9/25)+7/75)mL² </span>
<span>So that means the angular acceleration is: </span>
<span>α = τ_net/I_total </span>
<span>= ((2.75(3/5−2/5)+1/10)(mgL)cosθ)/((2.75(4... </span>
<span>The "m" cancels out. You're given "L" and "θ" and you know "g", so do the math (and don't forget to use consistent units).</span>