Answer:
Diagram C
Explanation:
We are given that Sulfur with atomic number 16.
We have to find that which diagram shows the electronic configuration of sulfur.
S=16
Its Diagram C
Answer:
Velocity of Pauli relative to Daniel = (-1.50ï + 3.90ĵ) m/s
x-component = -1.50 m/s
y-component = 3.90 m/s
Explanation:
Relative velocity of a body A relative to another body B, Vab, is given as
Vab = Va - Vb
where
Va = Relative velocity of body A with respect to another third body or frame of reference C
Vb = Relative velocity of body B with respect to that same third body or frame of reference C.
So, relative velocity can be given further as
Vab = Vac - Vbc
Velocity of Newton relative to Daniel = Vnd = 3.90 m/s due north = (3.90ĵ) m/s in vector form.
Velocity of Newton relative to Pauli = Vnp = 1.50 m/s due East = (1.50î) m/s in vector form
What is Pauli's velocity relative to Daniel?
Vpd = Vp - Vd
(Pauli's velocity relative to Daniel) = (Pauli's velocity relative to Newton) - (Daniel's velocity relative to Newton)
Vpd = Vpn - Vdn
Vpn = -Vnp = -(1.50î) m/s
Vdn = -Vnd = -(3.90ĵ) m/s
Vpd = -1.50î - (-3.90ĵ)
Velocity of Pauli relative to Daniel = (-1.50ï + 3.90ĵ) m/s
Hope this Helps!!!!
Answer:
The statement is true: velocity and acceleration have opposite directions in the interval of braking.
Explanation:
Let's say we have a velocity
.
The acceleration
is the rate of change of the velocity
. This means that if
is <em>increasing during</em> time, then
must be positive. But if
is <em>decreasing over</em> time, then
will be negative (even though the velocity is positive).
Mathematically:

decreases ⇒
⇒
.
Example:

Answer:
the maximum is I₁ axis of rotation at the end
the minimum moment is I₂ axis of rotation at the center of mass
Explanation:
For this exercise we use the definition moment of inertia
I = ∫ r² dm
for bodies of high symmetry it is tabulated; In this case we can approximate a broomstick to a thin rod, the moment of inertia with respect to a perpendicular axis when varying are
at one end
I₁ = ⅓ mL²
in in center
I₂ =
m L²
There is another possible axis of rotation around the axis of the broom, in this case we have a solid cylinder
I₃ =
m r²
remember that the diameter of the broom is much smaller than its length, therefore this moment of inertia is very small
when examining the different moments of inertia:
the maximum is I₁ axis of rotation at the end
the minimum moment is I₂ axis of rotation at the center of mass