The potential difference does work on the electron. The work is given by:
W = Vq
W = work, V = potential difference, q = electron charge
Given values:
V = 6V, q = 1.6x10^-19C
Plug in and solve for W:
W = 6(1.6x10^-19)
W = 0.96aJ
4 to 1
Explanation If you look at it as a fraction 160/40 and you reduce it down as far as you can you get 4/1
Answer:
The right response will be "450 volts".
Explanation:
The given values are:
R1 = 4.00 cm
R2 = 6.00 cm
q1 = +6.00 nC
q2 = −9.00 nC
As we know,
The potential difference between the two shell's difference will be:
⇒ ![\Delta V=K[(\frac{q1}{R1}+\frac{q2}{R2})-(\frac{q1}{R1} +(\frac{q2}{R2}))]](https://tex.z-dn.net/?f=%5CDelta%20V%3DK%5B%28%5Cfrac%7Bq1%7D%7BR1%7D%2B%5Cfrac%7Bq2%7D%7BR2%7D%29-%28%5Cfrac%7Bq1%7D%7BR1%7D%20%2B%28%5Cfrac%7Bq2%7D%7BR2%7D%29%29%5D)
![=K[\frac{q1}{R2}-\frac{q1}{R1} ]](https://tex.z-dn.net/?f=%3DK%5B%5Cfrac%7Bq1%7D%7BR2%7D-%5Cfrac%7Bq1%7D%7BR1%7D%20%5D)
On substituting the values, we get
Δ 
It would be 12,000 because newton’s third 2nd law states F=ma (force=matter x acceleration) so 30x400 would be your force .
please mark brainliest and i hope this helps!
For an ideal transformer power loss is assumed to be zero
i.e. the power in primary coil due to input voltage must be equal to power in secondary coil due to output voltage
this can be written in form of equation

here we know that


![i_1 = 10 A{/tex]now we will use above equation[tex]140*3.5 = 10 * V_1](https://tex.z-dn.net/?f=i_1%20%3D%2010%20A%7B%2Ftex%5D%3C%2Fp%3E%3Cp%3Enow%20we%20will%20use%20above%20equation%3C%2Fp%3E%3Cp%3E%5Btex%5D140%2A3.5%20%3D%2010%20%2A%20V_1)

So primary coil voltage is 49 Volts