The answers are the following:
1. encouraging them to act and unite
2. valiant and courageous
Answer:
the spring coefficient is
k=16N/m
Explanation:
Hooks law states that provided the elasticity of a material is not exceeded the extension e is proportional to the applied force
Step one
Analysis of the problem
From analysis of the problem
The mass has a potential energy due to the height it was dropped from, the potential energy is then stored in the spring since it was dropped on the spring which compresses it by 0.5m
Step two
Data
Mass of object m=0.2kg
Height of building =10m
Compression of spring e=0.5m
Spring constant k=?
Step three
According to the principle of energy conservation
mgh=1/2(k*e^2)
Making k subject of formula we have
k=2mgh/e^2
Substituting our data into the expression to get k
Assuming g=9.81m/s
k=2*0.2*10/0.5^2
k=4/0.25
k=16N/m
Answer : The magnitude of the orbital angular momentum for its most energetic electron is,
Explanation :
The formula used for orbital angular momentum is:
where,
L = orbital angular momentum
l = Azimuthal quantum number
As we are given the electronic configuration of Fe is,
Its most energetic electron will be for 3d electrons.
The value of azimuthal quantum number(l) of d orbital is, 2
That means, l = 2
Now put all the given values in the above formula, we get:
Therefore, the magnitude of the orbital angular momentum for its most energetic electron is,
Answer:
a) w = 25.1 rad/s, b) θ = 0.9599 rad
, c) α = 328.1 rad/s² d) t= 0.0765 s
Explanation: Let's work on this exercise with the equations of angular kinematics
a) The angular velocity is
w = 4.00 rev / s (2π rad / 1 rev)
w = 25.1 rad/s
b) let's reduce the angle of degrees to radians
θ = 55 ° (π rad / 180 °)
θ = 0.9599 rad
c) Let's use the initial angular velocity as the system part of the rest is zero
w² = w₀² + 2 α θ
α = w² / 2 θ
α = 25.1²/2 0.9599
α = 328.1 rad / s²
d)
w = w₀ + α t
t = w / α
t = 25.1 / 328.1
t= 0.0765 s