Explanation:
1. Mass of the proton, 
Wavelength, 
We need to find the potential difference. The relationship between potential difference and wavelength is given by :



V = 45.83 volts
2. Mass of the electron, 
Wavelength, 
We need to find the potential difference. The relationship between potential difference and wavelength is given by :




V = 84109.27 volt
Hence, this is the required solution.
<span>If there isn't any force then the normal contact force will be
N=m*g=7.5*9.81=73.58N
which is 73.58-23=50.58N less
so, there the person must pull at 23 degree upward
break down the tension in two components, vertical and horizontal.
vertical tension= 50.58=T*sin23
T=50.58/sin23=129.45N</span>
Answer:
= ( ρ_fluid g A) y
Explanation:
This exercise can be solved in two parts, the first finding the equilibrium force and the second finding the oscillating force
for the first part, let's write Newton's equilibrium equation
B₀ - W = 0
B₀ = W
ρ_fluid g V_fluid = W
the volume of the fluid is the area of the cube times the height it is submerged
V_fluid = A y
For the second part, the body introduces a quantity and below this equilibrium point, the equation is
B - W = m a
ρ_fluid g A (y₀ + y) - W = m a
ρ_fluid g A y + (ρ_fluid g A y₀ -W) = m a
ρ_fluid g A y + (B₀-W) = ma
the part in parentheses is zero since it is the force when it is in equilibrium
ρ_fluid g A y = m a
this equation the net force is
= ( ρ_fluid g A) y
we can see that this force varies linearly the distance and measured from the equilibrium position
Answer: angular displacement in rad = 3038.45 rad
angular displacement in rev = 483.589 rev
Explanation: mathematically
Angular velocity = angular displacement / time taken.
Angular velocity = 33.5 rad/s, time taken = 90.7s
33.5 = angular displacement /90.7
Angular displacement = 33.5 * 90.7 = 3038.45 rad
But 1 rev =2π
Hence 3038.45 rad to rev is
3038.45/2π = 483.599 rev
Answer:
the first one is Primary
the second one I think it's Mature but I don't know