In Case 1, a force F is pushing perpendicular on an object a distance L/2 from the rotation axis. In Case 2 the same force is pushing at an angle of 30 degrees a distance L from the axis. In which case is the torque due to the force about the rotation axis biggest?

Your answer is 1.5 hope i  

Answerthe energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.explanationwe define electric potential as the ratio electric potential energy to the charge,    then the following gives us the potential difference,      the formula  may be on your formula sheet.the change in electric potential energy would be the energy that would needed to move it from point a to point b. solving for that, we get    using q = 8.00  × 10⁻¹⁹ c and  δv = 2.00 v, we get    the energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.

Accelerating forwards; fa > ff            fn                           |ff< · > fa                      |                     fgfn = fg. so ∑f is just equal to sum of fa and ff (fn and fg cancel)find the distance the pickle travels. v₀ = 0 m/sd = 6.0 mt = 4.0 suse kinematic equationd = v₀t + at²/2d = 0t + at²/2d = at²/2a = 2d/t²a = 2(6.0 m)/(4.0 s)² a = 0.75 m/s² forwards∑f = fa + ffwe are given that ff = 3.8 n backwardsfa = ∑f - fffa = ma - ff fa = (25.0 kg)(0.75 m/s²) - (-3.8n)fa = 22.55 n forwardssow = fa · dw = (22.55 n)(6.0 m)w = 135 jw = 140 jthe work done to move the object is 140 j.

Anyone knows the answer before i take my exam.