I think it's in the same column as a sodium on the periodic table because chemical properties are based on their electron structure and the columns called groups are elements with the same shell structure which increases in mass as you go down the column, therefore the elements which are similar to sodium must be in the same group.
Is this a question or a statement. This statement would be incorrect. <span />
Kinetic energy. It decreases as you move further away from the nucleus. Electrons further away from the nucleus are held more weakly by the nucleus and so can be removed by spending less energy which is why these electrons will have higher energy
Speed would such a block have if pushed horizontally 106 m along a frictionless track by such a laser is 0.127 m / s
First, it is necessary to find the radiation pressure on the surface. You will find it using the following formula:
P = P / (πr ^ 2) c
where P is the pressure and c is the speed of light in vacuum
P = 27 * 10 ^ 6 / π (0.2 / 2) ^ 2 * (3 * 10 ^ 8)
= 286.62× = 2866N / m ^ 2.
Then you must calculate the force (F) and the acceleration (a). This is done through the formulas:
F = P * (πr ^ 2)
F = 2866 * π * (0.2 / 2) ^ 2 = 0.089N
As, a = F / m
a = 0.089 / 104 = 0.00085m / s ^ 2
You can now calculate the speed.
V = √2ad
V = √2 *0.00085 * 106
V = 0.127 m / s
The complete question is: You've recently read about a chemical laser that generates a 20.0-cm-diameter, 27.0 MW laser beam. One day, after physics class, you start to wonder if you could use the radiation pressure from this laser beam to launch small payloads into orbit. To see if this might be feasible, you do a quick calculation of the acceleration of a 20.0-cm-diameter, 104 kg, perfectly absorbing block. What speed would such a block have if pushed horizontally 106 m along a frictionless track by such a laser? Express your answer with the appropriate units.
We will use formula for the orbital velocity of Venus, which is v = 35.02 km/s.
An average distance to the Sun ( In kilometers ) is:
R = 0.723 * 149,579,871 km= 108,150,260 km.
Than we will calculate the orbital period ( T ).
v = 2 π R / T
T = 2 π R / v
T = 2 * 3.14 * 108,150,260 km / 126,072 km/s
T = 5389.75 s ≈ <span>224.5 days
The orbital period of Venus is approximately 224.5 days.</span>