Answer:
E(final)/E(initial)=2
Explanation:
Applying the law of gauss to two parallel plates with charge density equal σ:
So, if the charge is doubled the Electric field is doubled too
E(final)/E(initial)=2
Answer:
B meet A 0.01 km east of flagpole
Explanation:
given data
distance A = 5.7 km west
velocity V1 = 8.9 km/h
distance B = 4.5 km east
velocity V2 = 7 km/h
to find out
How far runners from the flagpole, when paths cross
solution
we know A and B are 5.7 + 4.5 = 10.2 km apart
and we consider here B will run distance x km for meet
so time will be for B is
time B = distance / velocity
time B = x / 7 ...................1
and
for A distance for meet = ( 10.2 - x ) km
so time A = distance / velocity
time A = ( 10.2 - x ) / 8.9 .............2
now equating equation 1 and 2
time A = time B
x / 7 = ( 10.2 - x ) / 8.9
x = 4.490
so distance of B run for meet is 4.490 km
so distance from the flagpole when their paths cross is 4.5 - 4.490 = 0.01 km
so B meet A 0.01 km east of flagpole
Answer: The electric repulsion between the two protons is stronger than the gravitational attraction.
Explanation: Please see the attachments below
A = (v - u) / t
a = acceleration
v = final velocity
u = initial velocity
t = time
45 = (v - 300) / 10
45 × 10 = v - 300
450 + 300 = v
v = 750 m/s
Hope this helps!
P.S. Let me know if you need an explanation