The circumference of a circle is (2π · the circle's radius).
The length of a semi-circle is (1π · the circle's radius) =
(π · 14.8) = 46.5 (rounded)
(The unit is the same as whatever the unit of the 14.8 is.)
Well, first of all, there's no such thing as "fully charged" for a capacitor.
A capacitor has a "maximum working voltage", because of mechanical
or chemical reasons, just like a car has a maximum safe speed. But
anywhere below that, cars and capacitors do their jobs just fine, without
any risk of failing.
So we have a capacitor that has some charge on it, and therefore some
voltage across it. From the list of choices above . . .
<span>-- Both plates have the same amount of charge.
Yes. And both plates have opposite TYPES of charge.
One plate is loaded with electrons and is negatively charged.
The other plate is missing electrons and is positively charged.
-- There is a potential difference between the plates.
Yes. That's the "voltage" mentioned earlier.
It's a measure of how badly the extra electrons want to jump
from the negative plate to the positive plate.
-- Electric potential energy is stored.
Yes. It's the energy that had to be put into the capacitor
to move electrons away from one plate and cram them
onto the other plate.
</span>
You're talking about a grain of sand or a stone or a rock that's drifting in space, and then the Earth happens to get in the way, so the stone falls down to Earth, and it makes a bright streak of light while it's falling through the atmosphere and burning up from the friction.
-- While it's drifting in space, it's a <em>meteoroid</em>.
-- While it's falling through the atmosphere burning up and making a bright streak of light, it's a <em>meteor</em>.
-- If it doesn't completely burn up and there's some of it left to fall on the ground, then the leftover piece on the ground is a <em>meteorite</em>.
Answer:
<h2>The angular velocity just after collision is given as</h2><h2>
</h2><h2>At the time of collision the hinge point will exert net external force on it so linear momentum is not conserved</h2>
Explanation:
As per given figure we know that there is no external torque about hinge point on the system of given mass
So here we will have
now we can say
so we will have
Linear momentum of the system is not conserved because at the time of collision the hinge point will exert net external force on the system of mass
So we can use angular momentum conservation about the hinge point
