Batteries supply electrons to the circuit by releasing negatively charged atoms or ions. These ions are produced by the batteries through a chemical reaction that spontaneously occurs within the battery. So the negative end of the battery pushes the ions towards the positive end of the circuit with the help of the voltage. This is why eventually, batteries "run out" when the electrode is used up and the chemical reaction can no longer continue.
<span>Well, since it's in the shape of a wheel and the person walks around the edge of it, they must have a centripetal acceleration. Since a=v^2/r you can solve for "v" using 2.20 as your "a" and 59.5 as your "r" (r=half of the diameter).
</span> a=v^2/r
v=(a*r)^(1/2)=((2.20)*(59.5))^(1/2)=<span>
<span>11.44 m/s.
</span></span><span> After you get "v," plugged that into T=2 pi r/ v. This will give you the 1rev per sec.
</span> T=2 pi r/ v= T=(2)*(pi)*(59.5)/(11.44)= <span>
<span>32.68 rev/s
</span></span> Use dimensional analysis to get rev per min (1rev / # sec) times (60 sec/min).
(32.68 rev/s)(60 s/min)=<span>
<span>1960.74 rev/min
</span></span>
Answer:
Explanation:
<u>LC Circuit</u>
It's a special circuit made of three basic elements: The AC source, a capacitor, and an inductor. The charge, current, and voltage are oscillating when there is an interaction between the electric and magnetic fields of the elements. The following variables will be used for the formulas:
= charge of the capacitor in any time 
= initial charge of the capacitor
=angular frequency of the circuit
= current through the circuit in any time
The charge in an LC circuit is given by
The current is the derivative of the charge
We are given
It means that
![q(t_1) = q_0 \, cos (\omega t_1 )=q_1\ .......[eq 1]](https://tex.z-dn.net/?f=q%28t_1%29%20%3D%20q_0%20%5C%2C%20cos%20%28%5Comega%20t_1%20%29%3Dq_1%5C%20.......%5Beq%201%5D)
![i(t_1) = - \omega q_0 \, sin(\omega t_1)=i_1.........[eq 2]](https://tex.z-dn.net/?f=i%28t_1%29%20%3D%20-%20%5Comega%20q_0%20%5C%2C%20sin%28%5Comega%20t_1%29%3Di_1.........%5Beq%202%5D)
From eq 1:
From eq 2:
Squaring and adding the last two equations, and knowing that
Operating
Solving for
Now we know the value of 
, we repeat the procedure of eq 1 and eq 2, but now at the second time 
, and solve for 
Solving for
Now we replace the given values. We'll assume that the placeholder is a pi for the frequency, i.e.
Finally
Answer:
1 W = 1 J / sec Definition of watt is 1 joule / sec
So if a bulb uses 75 J / sec it must use
75 J/s * 60 sec / min = 4500 J/min energy used by bulb
If bulb is 15% efficient then the light delivered is
P = 4500 J / min * .15 = 675 J / min
