It involves electrons.
The cathode is the electrode where electron deficient ions move to.
While the anode is electrode where electron excess ions move to.
So the relationship between Cathode and Anode involves electrons.
C.
So the area under a velocity time graph is distance or displacement, if you have done calculus yet you will understand that if you take the integral of a velocity function then you end up with displacement. Thats for later understanding however.
So this appears to be a right triangle so we can find the area of a triangle as:
0.5bh = A
Since our area is 10 meters lets alter our formula a bit to fit the situation:
Our base here is time and our height is velocity so:
0.5tv = Δx
So we can read off the graph that our velocity at the end, or our final velocity appears to be near 2.0 m/s
So we have v, and Δx so lets isolate for time by dividing by v and 0.5
t = Δx / 0.5v
Now lets plug all that in:
t = 10 / 0.5(2)
t = 10 seconds
Hope this helped!
Answer:
Moving a unit "positive" test charge from A to B will result in a reduction in potential
V = K Q / R potential at a point
V2 - V1 = K Q (1 / .4 - 1 / .15) = = k Q (.15 - .4) / .06 = -4.17 K Q
V2 - V1 = -4.17 * 9 & 10E9 * 6.25 E-8
V2 - V1 = -4.17 * 562.5 J/C
V = - 2346 Volts
Answer: C. electricity and magnetism
Explanation:
