Answer:
ω' = 2.5 rad/s
Explanation:
mass of cockroach, m = 4 kg
mass of disk, M = 6 kg
Radius of disc= R
initial angular velocity, ω = 2 rad/s
Let the final angular velocity is ω'
As no external torque is applied, so the angular momentum is constant.
Angular momentum = Moment of inertia x angular velocity
I ω = I' ω'
ω' = 2.5 rad/s
Well, Godess, that's not a simple question, and it doesn't have
a simple answer.
When the switch is closed . . .
"Conventional current" flows out of the ' + ' of the battery, through R₁ ,
then through R₂ , then through R₃ . It piles up on the right-hand side of
the capacitor (C). It repels the ' + ' charges on the left side of 'C', and
those flow into the ' - ' side of the battery. So the flow of current through
this series circuit is completely clockwise, around toward the right.
That's the way the first experimenters pictured it, that's the way we still
handle it on paper, and that's the way our ammeters display it.
BUT . . .
About 100 years after we thought that we completely understand electricity,
we discovered that the little tiny things that really move through a wire, and
really carry the electric charge, are the electrons, and they carry NEGATIVE
charge. This turned our whole picture upside down.
But we never changed the picture ! We still do all of our work in terms of
'conventional current'. But the PHYSICAL current ... the actual motion of
charge in the wire ... is all exactly the other way around.
In your drawing ... When the switch is closed, electrons flow out of the
' - ' terminal on the bottom of the battery, and pile up on the left plate of
the 'C'. They repel electrons off of the right-side of 'C', and those then
flow through R₃ , then through R₂ , then through R₁ , and finally into the
' + ' terminal on top of the battery.
Those are the directions of 'conventional' current and 'physical' current
in all circuits.
In the circuit of YOUR picture that you attached, there's more to the story:
Battery current can't flow through a capacitor. Current flows only until
charges are piled up on the two sides of 'C' facing each other, and then
it stops.
Wait a few seconds after you close the switch in the picture, and there is
no longer any current in the loop.
To be very specific and technical about it . . .
-- The instant you close the switch, the current is
(battery voltage) / (R₁ + R₂ + R₃) amperes
but it immediately starts to decrease.
-- Every (C)/((R₁ + R₂ + R₃) seconds after that, the current is
e⁻¹ = about 36.8 %
less than it was that same amount of time ago.
Now, are you glad you asked ?
The solution for this problem is computed by through this formula, F = kQq / d²Plugging in the given values above, we can now compute for the answer.
F = 8.98755e9N·m²/C² * -(7e-6C)² / (0.03m)² = -489N, the negative sign denotes attraction.
Yes, that's why the atom is neutral. A proton carries a positive charge,
and an electron carries a negative charge. When their numbers are
equal, the atom has no 'net' charge, and it's a 'neutral' atom.
The force applied to the second ball by the first ball is 6.734 × 10^-4 N.
<h3>What is impulse of force?</h3>
The impulse of force is defined as the sum of the average force and the duration it is applied.
If the mass of the item remains constant, the impulse of force equals the change in momentum of the object.
Given that: mass of a metal sphere: m = 0.026 kg.
Initial speed of the sphere: u = 3.7 m/s.
When the sphere stops completely, its change in momentum = mu - 0
= 0.026×3.7 N-s.
= 0.0962 N-s.
As the spheres are in contact for 0.007s before the second sphere is shot off down the track, the force applied to the second ball =
change in momentum of 1st ball × time of contact
= 0.0962 × 0.007 N
= 0.0006734 N
= 6.734 × 10^-4 N.
Hence, the force applied to the second ball is 6.734 × 10^-4 N.
Learn more about impulse force here:
brainly.com/question/29787329
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