V = I * R
Where V is the voltage, I is the current and R is the resistance. Using Ohm's law, you require resistance to find the current through the wire. Technically, if the wire has a resistance of 0, you will get infinite current. But this isn't possible. Maybe the negligible resistance refers to the battery's internal resistance - not the wire's resistance.
Although many characteristics are common<span> throughout the </span>group<span>, the heavier metals such as Ca, Sr, Ba, and Ra are almost as reactive as the </span>Group<span> 1 Alkali Metals. All the </span>elements<span> in </span>Group 2 have two<span> electrons in their valence shells, giving them an oxidation state of +</span><span>2.</span>
If the container explodes there is no pressure, becuase all your gas has escaped its container, there for, you ain’t got no gas
I can't see that cube from here.
But if the length of the side of the cube is ' K ' units,
then the surface area of the cube is 6K² units², and
the volume of the cube is K³ units³.
The ratio of the surface area to the volume is
(6K² units²) / (K³ units³) = (6) / (K units) .
So for example, if the side of the cube is 2 inches, then
the ratio of surface area to volume is "3 per inch".
That's the answer. I did the whole thing in order to earn
the points, but I don't expect you to understand much of it,
because I see from your username that you suck at math.
I'm sorry you decided that. Now that you've put up the
brick wall, it'll be even harder for any math to find its way
in there, and you'll miss out on a lot of the fun.
Answer:
L/2
Explanation:
Neglect any air or other resistant, for the ball can wrap its string around the bar, it must rotate a full circle around the bar. This means the ball should be able to swing to the top position where it's directly above the bar. By the law of energy conservation, this happens when the ball is at the same level as where it's previously released vertically. It means the swinging radius around the bar must be at least half of the string length.
So the distance d between the bar and the pivot should be at least L/2
