The scientist who suggested that atoms contain a hard core containing most of the mass of an atom was Ernest Rutherford.
Answer: On Monday, XX/XX, the XXXX you presented in our weekly review were not accurate and I’m worried that this reflected poorly on our team’s credibility. This could negatively impact senior leadership’s confidence in our team’s effectiveness at a time when headcount is already being scrutinized and potentially eliminated. Next time you present XXXX, my expectation is that everything will have been double or triple-checked in advance of the meeting so that all figures are reported accurately.
Answer:
The potential energy is transformed into kinetic energy
Explanation:
This particular case is defined as the principle of energy conservation since energy is not created or destroyed only transforms. When you have potential energy it can be transformed into kinetic energy or vice versa. In this problem, we have the case of a ball that sits on a desk and then falls to the ground. In this way the ground will be taken as a reference point, this is a point at which the potential energy will be equal to zero in such a way that when the ball is on the desktop that is above the reference line its potential energy will be maximum. As the ball drops its potential energy decreases, as the height relative to the ground (reference point) decreases. In contrast its kinetic energy increases and increases as it approaches the ground. So when it hits the ground it will have maximum kinetic energy and will be equal to the potential energy for when the ball was on the desk.
Therefore:
![E_{p} = potential energy [J] = E_{k} = kinetic energy [J]where:\\E_{p} =m*g*h\\m =mass [kg]\\g=gravity[m/s^2]\\h=elevation[m]\\E_{k} = \frac{1}{2} *m*v^{2} \\where:\\v=velocity [m/s]\\\frac{1}{2} *m*v^{2} = m*g*h](https://tex.z-dn.net/?f=E_%7Bp%7D%20%3D%20potential%20energy%20%5BJ%5D%20%3D%20E_%7Bk%7D%20%3D%20kinetic%20energy%20%5BJ%5Dwhere%3A%5C%5CE_%7Bp%7D%20%3Dm%2Ag%2Ah%5C%5Cm%20%3Dmass%20%5Bkg%5D%5C%5Cg%3Dgravity%5Bm%2Fs%5E2%5D%5C%5Ch%3Delevation%5Bm%5D%5C%5CE_%7Bk%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%2Am%2Av%5E%7B2%7D%20%5C%5Cwhere%3A%5C%5Cv%3Dvelocity%20%5Bm%2Fs%5D%5C%5C%5Cfrac%7B1%7D%7B2%7D%20%20%2Am%2Av%5E%7B2%7D%20%3D%20m%2Ag%2Ah)
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