Answer:

Explanation:
The gravitational force between the proton and the electron is given by

where
G is the gravitational constant
is the proton mass
is the electron mass
r = 3 m is the distance between the proton and the electron
Substituting numbers into the equation,

The electrical force between the proton and the electron is given by

where
k is the Coulomb constant
is the elementary charge (charge of the proton and of the electron)
r = 3 m is the distance between the proton and the electron
Substituting numbers into the equation,

So, the ratio of the electrical force to the gravitational force is

So, we see that the electrical force is much larger than the gravitational force.
Answer: 500 s
Explanation:
Speed
is defined as a relation between the distance
and time
:

Where:
is the speed of light in vacuum
is the distance between the Earth and Sun
is the time it takes to the light to travel the distance
Isolating
:


Finally:

Answer: a) 274.34 nm; b) 1.74 eV c) 1.74 V
Explanation: In order to solve this problem we have to consider the energy balance for the photoelectric effect on tungsten:
h*ν = Ek+W ; where h is the Planck constant, ek the kinetic energy of electrons and W the work funcion of the metal catode.
In order to calculate the cutoff wavelength we have to consider that Ek=0
in this case h*ν=W
(h*c)/λ=4.52 eV
λ= (h*c)/4.52 eV
λ= (1240 eV*nm)/(4.52 eV)=274.34 nm
From this h*ν = Ek+W; we can calculate the kinetic energy for a radiation wavelength of 198 nm
then we have
(h*c)/(λ)-W= Ek
Ek=(1240 eV*nm)/(198 nm)-4.52 eV=1.74 eV
Finally, if we want to stop these electrons we have to applied a stop potental equal to 1.74 V . At this potential the photo-current drop to zero. This potential is lower to the catode, so this acts to slow down the ejected electrons from the catode.
To solve the problem it is necessary to use Newton's second law and statistical equilibrium equations.
According to Newton's second law we have to

where,
m= mass
g = gravitational acceleration
For the balance to break, there must be a mass M located at the right end.
We will define the mass m as the mass of the body, located in an equidistant center of the corners equal to 4m.
In this way, applying the static equilibrium equations, we have to sum up torques at point B,

Regarding the forces we have,

Re-arrange to find M,



Therefore the maximum additional mass you could place on the right hand end of the plank and have the plank still be at rest is 16.67Kg
Im pretty sure its C- Wavelength