By how much would its speed reading increase with each second of fall? ... Ex 3.24 For a freely falling object dropped from rest, what is its acceleration at the end of the 5th second ... Pb 3.3 A ball is thrown straight up with an initial speed of 30 m/s. How high does it go, and how long is it in the air (neglecting air resistance)?.
Answer:
Simply drop the egg from one inch above your foot and it will not break.
Explanation:
Answer:
0.247 J = 247 mJ
Explanation:
From the principle of conservation of energy, the workdone by the applied force, W = kinetic energy change + electric potential energy change.
So, W = ΔK + ΔU =1/2m(v₂² - v₁²) + q(V₂ - V₁) where m = mass of particle = 5.4 × 10⁻² kg, q = charge of particle = 5.10 × 10⁻⁵ C, v₁ = initial speed of particle = 2.00 m/s, v₂ = final speed of particle = 3.00 m/s, V₁ = potential at surface A = 5650 V, V₂ = potential at surface B = 7850 V.
So, W = ΔK + ΔU =1/2m(v₂² - v₁²) + q(V₂ - V₁)
= 1/2 × 5.4 × 10⁻²kg × ((3m/s)² - (2 m/s)²) + 5.10 × 10⁻⁵ C(7850 - 5650)
= 0.135 J + 0.11220 J
= 0.2472 J
≅ 0.247 J = 247 mJ
A pendulum is probably the most common showing of this example. As the pendulum swings down, it converts its potential energy (height) into kinetic energy (velocity). At the lowest point the kinetic energy is the highest and the potential is the lowest. At the highest point in its swing the velocity is zero so the kinetic energy is zero and the potential energy is at a maximum (greatest height).
