This question is checking to see whether you understand the meaning
of "displacement".
Displacement is a vector:
-- Its magnitude (size) is the distance between the start-point and
the end-point, no matter what route might have been followed along
the way.
-- Its direction is the direction from the start-point to the end-point.
Talking about the Earth's orbit around the sun, we can forget about
the direction of the displacement, and just talk about its magnitude
(size).
If we pretend that the sun is not moving and dragging the whole
solar system along with it, then what do we see the Earth doing
in one year ?
We mark the place where the Earth is at the stroke of midnight
on New Year's Eve. Then we watch it as it swings around through
this gigantic orbit, all the way around the sun, and in a year, it's back
to the same point that we marked !
So what's the magnitude of the displacement in exactly one year ?
It's the distance between the start-point and the end-point. But the
Earth came back to the same place it started from, so there's no
separation at all between the start-point and the end-point.
The Earth covered a huge distance in that year, but the displacement
is zero.
(c) the fraction of energy lost during the impact
Answer:
d = 4.5079mm
Explanation:
from the question we are given that
= 0.082N/m
I₁= 30.6 A , I₂ = 60.4 A , and μ₀ = 4 π × 10⁻⁷ T · m / A
In order for the system to be in equilibrium, the repulsive magnetic force
per unit length on the top wire must equal the weight per unit length of the wire. Then we have
F/L = μ₀I₁I₂ / 2πd , making d as the subject of formular
we have that d = Lμ₀I₁I₂ / 2πF
= 4 π × 10⁻⁷ × 30.6 ×60.4 / 2π × 0.082
=45079.02× 10⁻⁷
in mm .. 45079.02× 10⁻⁷ × 10³
45079.02 ×10⁻⁴
d = 4.5079mm
Answer:
Part a)
Part b)
Explanation:
As we know that it makes half revolution in given time interval
so we have
now the angular speed is given as
now linear speed is given as
now we have
Now centripetal acceleration is given as
Part b)
Average acceleration of the cat is given as
