Answer:
If one cup falls down then there will be 59 cups left.
Answer:
0.363999909622
Explanation:
F = Force
m = Mass = 15.6 g
C = Drag coefficient
ρ = Density of air = 1.21 kg/m³
A = Surface area = 
v = Terminal velocity = 
s = Displacement = 150 m

Force is given by
F = ma

The drag coefficient is 0.363999909622 (ignoring negative sign)
Answer:
the final kinetic energy is 0.9eV
Explanation:
To find the kinetic energy of the electron just after the collision with hydrogen atoms you take into account that the energy of the electron in the hydrogen atoms are given by the expression:

you can assume that the shot electron excites the electron of the hydrogen atom to the first excited state, that is
![E_{n_2-n_1}=-13.6eV[\frac{1}{n_2^2}-\frac{1}{n_1^2}]\\\\E_{2-1}=-13.6eV[\frac{1}{2^2}-\frac{1}{1}]=-10.2eV](https://tex.z-dn.net/?f=E_%7Bn_2-n_1%7D%3D-13.6eV%5B%5Cfrac%7B1%7D%7Bn_2%5E2%7D-%5Cfrac%7B1%7D%7Bn_1%5E2%7D%5D%5C%5C%5C%5CE_%7B2-1%7D%3D-13.6eV%5B%5Cfrac%7B1%7D%7B2%5E2%7D-%5Cfrac%7B1%7D%7B1%7D%5D%3D-10.2eV)
-10.2eV is the energy that the shot electron losses in the excitation of the electron of the hydrogen atom. Hence, the final kinetic energy of the shot electron after it has given -10.2eV of its energy is:

Answer:

Explanation:
According to the Coulomb's law, the magnitude of the electrostatic force between two static point charges
and
, separated by a distance
, is given by

where k is the Coulomb's constant.
Initially,

The negative sign is taken with force F because the force is attractive.
Therefore, the initial electrostatic force between the charges is given by

Now, the objects are then brought into contact, so the net charge is shared equally, and then they are returned to their initial positions.
The force is now repulsive, therefore, 
The new charges on the two objects are

The new force is given by

Using (1),



Using (1),
When
,

When
,

Since, 
Therefore, 