Answer:
E1 = 10.15 * 10^4 N/C
E2 = 0
E3 = 10.15 *10^4 N/C
Explanation:
Given data:
Two 13 cm-long thin glass rods ( L ) = 0.13 m
charge (Q) = +11nC
distance between thin glass rods = 4 cm .
<u>Calculate the electric field strengths </u>
electric charge due to a single glass rod in the question ( E ) = 
equation 1 can be used to determine E1, E2 and E3 because the points lie within the two rods hence the net electric field produced will be equal to the difference in electric fields produced
applying equation 1 to determine E1
E1 =
( distance from 1 rod is 0.01 m and from the other rod is 0.03 )
= 
= 10.15 * 10^4 N/C
applying equation 1 to determine E2
E2 = 

therefore E2 = 0
E1 = E3
hence E3 = 10.15*10^4 N/C
Answer:
(a) 
(b) 
Explanation:
<u>Given:</u>
= The first temperature of air inside the tire = 
= The second temperature of air inside the tire = 
= The third temperature of air inside the tire = 
= The first volume of air inside the tire
= The second volume of air inside the tire = 
= The third volume of air inside the tire = 
= The first pressure of air inside the tire = 
<u>Assume:</u>
= The second pressure of air inside the tire
= The third pressure of air inside the tire- n = number of moles of air
Since the amount pof air inside the tire remains the same, this means the number of moles of air in the tire will remain constant.
Using ideal gas equation, we have

Part (a):
Using the above equation for this part of compression in the air, we have

Hence, the pressure in the tire after the compression is
.
Part (b):
Again using the equation for this part for the air, we have

Hence, the pressure in the tire after the car i driven at high speed is
.
<u>Disaster</u>:- a calamitous event, especially one occurring suddenly and causing great loss of life, damage, or hardship, as a flood / airplane crash.
:))
<span>Kinetic energy because it is taking the students to school.</span>
Gravitational potential energy, relative to some level =
(mass of the object)
times
(height above the reference level)
times
(acceleration due to gravity) .