Answer:
A) Emin = eV
B) Vo = (E_light - Φ) ÷ e
Explanation:
A)
Energy of electron is the product of electron charge and the applied potential difference.
The energy of an electron in this electric field with potential difference V will be eV. Since this is the least energy that the electron must reach to break out, then the minimum energy required by this electron will be;
Emin = eV
B)
The maximum stopping potential energy is eVo,
The energy of the electron due to the light is E_light.
If the minimum energy electron must posses is Φ, then the minimum energy electron must have to reach the detectors will be equal to the energy of the light minus the maximum stopping potential energy
Φ = E_light - eVo
Therefore,
eVo = E_light - Φ
Vo = (E_light - Φ) ÷ e
Answer:
n = 756.25 giga electrons
Explanation:
It is given that,
If the charge on the negative plate of the capacitor, 
Let n is the number of excess electrons are on that plate. Using the quantization of charges, the total charge on the negative plate is given by :
e is the charge on electron
or
n = 756.25 giga electrons
So, there are 756.25 giga electrons are on the plate. Hence, this is the required solution.
To solve this problem, we must remember about the law of
conservation of momentum. The initial momentum mist be equal to the final
momentum, that is:
m1 v1 + m2 v2 = (m1 + m2) v’
where v’ is the speed of impact
Since we are not given the masses of each car m1 and m2,
so let us assume that they are equal, such that:
m1 = m2 = m
Which makes the equation:
m v1 + m v2 = (2 m) v’
Cancelling m and substituting the v values:
50 + 48 = 2 v’
2 v’ = 98
v ‘ = 49 km/h
<span>The speed of impact is 49 km/h.</span>
Answer: In your right wrist
Explanation:
