Answer:
Magnitude of its angular momentum = 0.0017 kgm²/s
Explanation:
Angular momentum, L = Iω
I is mass moment of inertia and ω is angular velocity.
Phonograph is in disc shape,
Radius = 0.5 x 48 = 24 cm = 0.24 m
Angular velocity, ω = 3.2 rad/s
Mass, M = 18 g = 0.018 kg
Substituting
Magnitude of its angular momentum = 0.0017 kgm²/s
You clearly identified the pole you're talking about as the
"north-seeking" pole. Assuming your integrity and sincerity,
we would then naturally expect that pole to seek north, and
point to Earth's north magnetic pole.
I'm confident in this answer also because I have several of
these devices hanging from the ceiling of my office, and I can
attest to the fact that on most clear days, they do in fact point
toward Earth's north magnetic pole.
Answer:
Part a)
Part b)
Explanation:
Part a)
Initial value of magnetic flux is given as
so we have
Final flux through the loop is given as
now EMF is given as
Part b)
If magnetic field is constant while Area is changing
So EMF is given as
Well, one is more effeciant than the other. I think it would run on less gass.
Answer:
ΔK.E = 2.5 × 10⁻³ J
Explanation:
Given data in the question, we have:
Charge of the particle, q = 5.0 μC = 5 × 10 ⁻⁶ C
Initial speed of the particle, v = 55 m/s
The potential difference, ΔV = 500 V
Now, the gain in kinetic energy is given as
ΔK.E = q × ΔV
on substituting the values in the above formula, we get
ΔK.E = 5 × 10 ⁻⁶ C × 500 V
or
ΔK.E = 2.5 × 10⁻³ J
