Answer:
84.82N/C.
Explanation:
The x-components of the electric field cancel; therefore, we only care about the y-components.
The y-component of the differential electric field at the center is
.
Now, let us call
the charge per unit length, then we know that
;
therefore,


Integrating

![$E = \frac{k \lambda }{R}*[-cos(\pi )+cos(0) ]$](https://tex.z-dn.net/?f=%24E%20%3D%20%5Cfrac%7Bk%20%5Clambda%20%20%20%7D%7BR%7D%2A%5B-cos%28%5Cpi%20%29%2Bcos%280%29%20%5D%24)

Now, we know that


and the radius of the semicircle is

therefore,


Answer:
meteor
Explanation:
A asteroid stays still and a meteor goes fast
Floating. When you have no gravity you have nothing to be pushing you down to the floor so that would be an example of no gravity pushing on you.
<h2>
Answer:</h2>
38.14Ω
<h2>
Explanation:</h2>
Let's solve this question using Ohm's law which states that the current (I) flowing through a conductor is directly proportional to the potential difference or voltage (V) across it. Mathematically;
V = I R -------------------(i)
<em>Where</em>;
R is the constant of proportionality called resistance of the conductor and is measured in Ohms (Ω)
<em>From the question;</em>
V = 18.5V
I = 0.485A
<em>Substitute these values into equation (i) as follows;</em>
18.5 = 0.485 x R
<em>Solve for R;</em>
R = 18.5 / 0.485
R = 38.14Ω
Therefore the resistance of the bulb is 38.14Ω
Answer:
Ecu/Eag = 0.46
Explanation:
E = PI/A
Ecu = Pcu × I/A
Pcu = 1.72×10^-8 ohm-meter
r = 0.8 mm = 0.8/1000 = 8×10^-4 m
A = πr^2 = π×(8×10^-4)^2 = 6.4×10^-7π
Ecu = 1.72×10^-8I/6.4×10^-7π = 0.026875I/1
Eag = Pag × I/A
Pag = 1.47×10^-8 ohm-meter
r = 0.5 mm = 0.5/1000 = 5×10^-4 m
A = πr^2 = π × (5×10^-4)^2 = 2.5×10^-7π
Eag = 1.47×10^-8I/2.5×10^-7π = 0.0588I/π
Ecu/Eag = 0.026875I/π × π/0.0588I = 0.46