Complete question:
A uniform electric field is created by two parallel plates separated by a
distance of 0.04 m. What is the magnitude of the electric field established
between the plates if the potential of the first plate is +40V and the second
one is -40V?
Answer:
The magnitude of the electric field established between the plates is 2,000 V/m
Explanation:
Given;
distance between two parallel plates, d = 0.04 m
potential between first and second plate, = +40V and -40V respectively
The magnitude of the electric field established between the plates is calculated as;
E = ΔV / d
where;
ΔV is change in potential between two parallel plates;
d is the distance between the plates
ΔV = V₁ -V₂
ΔV = 40 - (-40)
ΔV = 40 + 40
ΔV = 80 V
E = ΔV / d
E = 80 / 0.04
E = 2,000 V/m
Therefore, the magnitude of the electric field established between the plates is 2,000 V/m
Answer:
a) 11 m/s
b) 0.0564 s
Explanation:
Given:
m = 2100 kg
vi = 22 ..... m/s before collision
vf = 0 ......after collision to stop
Δs = 0.62 distance traveled after collision .. crumpling of truck
Part a
Part b
TP hydrolysis distinct from any incorporation into the chain.
If you’re talking about atoms, I hope this helps
Atoms consist of three basic particles: protons, electrons, and neutrons. The nucleus (center) of the atom contains the protons (positively charged) and the neutrons (no charge). The outermost regions of the atom are called electron shells and contain the electrons (negatively charged)
