Answer:


Explanation:
m = Mass of each the cars = 
= Initial velocity of first car = 3.46 m/s
= Initial velocity of the other two cars = 1.4 m/s
v = Velocity of combined mass
As the momentum is conserved in the system we have

Speed of the three coupled cars after the collision is
.
As energy in the system is conserved we have

The kinetic energy lost during the collision is
.
1. Amperes, is the SI unit (also a fundamental unit) responsible for current.
2.
Δq over Δt technically
Rearrange for Δq
I x Δt = Δq
1.5mA x 5 = Δq
Δq = 0.0075
Divide this by the fundamental charge "e"
Electrons: 0.0075 / 1.60 x 10^-19
Electrons: 4.6875 x 10^16 or 4.7 x 10^16
3. So we know that the end resistances will be equal so:
ρ = RA/L
ρL = RA
ρL/A = R
Now we can set up two equations one for the resistance of the aluminum bar and one for the copper: Where 1 represents aluminum and 2 represents copper

We are looking for L2 so we can isolate using algebra to get:

If you fill in those values you get 0.0205
or 2.05 cm
Answer: B
Explanation: i learned it last year
Answer:

Explanation:
= Permittivity of free space = 
A = Area
h = Altitude = 600 m
Electric flux through the top would be
(negative as the electric field is going into the volume)
At the bottom

Total flux through the volume

Electric flux is given by

Charge per volume is given by

The volume charge density is 
no it doesn't why because I think that it is not the same but different.