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
The work done by tension force of 14N applied on the laptop by a rope as it moves 2.0 mm up the slope is 0.028 J
W = F d cos θ
W = Work done
F = Force
d = Displacement
θ = Angle between force and displacement vector
F = 14 N
d = 2 mm = 0.002 m
θ = 0
W = 14 * 0.002 * 1
W = 0.028 J
Work done is the change in energy of an object. So if an object moves a certain distance, work is done on the object. If the force and displacement are perpendicular to each other there is no work done on the object.
Therefore, the work done by tension on the laptop is 0.028 J
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Answer:
b) q large and m small
Explanation:
q is large and m is small
We'll express it as :
q > m
As we know the formula:
F = Eq
And we also know that :
F = Bqv
F = 
Bqv =
or Eq =
Assume that you want a velocity selector that will allow particles of velocity v⃗ to pass straight through without deflection while also providing the best possible velocity resolution. You set the electric and magnetic fields to select the velocity v⃗ . To obtain the best possible velocity resolution (the narrowest distribution of velocities of the transmitted particles) you would want to use particles with q large and m small.
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