Sometimes arithmetic problems can be solved much more easily using the dimensional analysis approach. You focus on the units of the given information. Then, you manipulate them applying the laws of algebra where like units cancel, in order to end up with the unit of the unknown.
Given:
-50 nc/step
31 steps
Unknown: charge
Thus,
Charge = -50 nc/step * 31 steps =<em> -1550 nc</em>
Answer:
I = 4.75 A
Explanation:
To find the current in the wire you use the following relation:
(1)
E: electric field E(t)=0.0004t2−0.0001t+0.0004
ρ: resistivity of the material = 2.75×10−8 ohm-meters
J: current density
The current density is also given by:
(2)
I: current
A: cross area of the wire = π(d/2)^2
d: diameter of the wire = 0.205 cm = 0.00205 m
You replace the equation (2) into the equation (1), and you solve for the current I:

Next, you replace for all variables:

hence, the current in the wire is 4.75A
Amanda might be suffering from a disease called Sickle Cell Anemia. It is an inherited red blood cell disorder.
Answer:
C.
Explanation:
A meter is 8.56 centimeters longer than a yard. Something to keep in mind is that a meter is about 10% longer than a yard.
Hope this helps :)
Answer:
Explanation:
Potential due to a charged metallic sphere having charge Q and radius r on its surface will be
v = k Q / r . On the surface and inside the metallic sphere , potential is the same . Outside the sphere , at a distance R from the centre potential is
v = k Q / R
a ) On the surface of the shell , potential due to positive charge is
V₁ = 
On the surface of the shell , potential due to negative charge is
V₁ = 
Total potential will be zero . they will cancel each other.
b ) On the surface of the sphere potential
= 
= 22.5 x 10⁵ V
On the surface of the sphere potential due to outer shell
= 
= -9 x 10⁵
Total potential
=( 22.5 - 9 ) x 10⁵
= 13.5 x 10⁵ V
c ) In the space between the two , potential will depend upon the distance of the point from the common centre .
d ) Inside the sphere , potential will be same as that on the surface that is
13.5 x 10⁵ V.
e ) Outside the shell , potential due to both positive and negative charge will cancel each other so it will be zero.