Answer:
Step-by-step explanation:
The work W required for the transfer is equal to the energy difference
![W = \Delta E = q \ \Delta V](https://img.qammunity.org/2020/formulas/physics/college/4s7qbcd3g6lel9td0vfj0rs3meweon31mv.png)
where q is the charge we want to transfer and ΔV is the potential difference.
So, the work needed will be
![W = 9,000 \ C * 220 V](https://img.qammunity.org/2020/formulas/physics/college/8luzuris5pta9mhhm5j5x4gq0qy981nhb2.png)
![W = 1,980,000 \ J](https://img.qammunity.org/2020/formulas/physics/college/k9a0vlsvzz1vro8l70p3ow8yrfkbjy4js2.png)
Now, the average power is:
![< power > = (work)/(time)](https://img.qammunity.org/2020/formulas/physics/college/7ic2vnirfarzgs55o55ahvvgbzne1qhuof.png)
As the Watt is Joule/second, we need the 45 min multiplied by 60:
![time = 45 \ min * 60 (s)/(min)](https://img.qammunity.org/2020/formulas/physics/college/7tyc37ydzliox7zdzj3w9noiapccemes71.png)
![time = 2,700 \ s](https://img.qammunity.org/2020/formulas/physics/college/zzfp541kaayccs1ha329xsqj8byum9fu96.png)
Taking all this together:
![< power > = (1,980,000 \ J)/(2,700 \ s)](https://img.qammunity.org/2020/formulas/physics/college/oxhrbnlx8527bct5oih26bhp0h7bmu972e.png)
![< power > = (1,980,000 \ J)/(2,700 \ s)](https://img.qammunity.org/2020/formulas/physics/college/oxhrbnlx8527bct5oih26bhp0h7bmu972e.png)
![< power > = 733.333 \ W](https://img.qammunity.org/2020/formulas/physics/college/d5xr23lwsp5s1ialk3bl7lk4lzhzzevgwx.png)
or
![< power > = 0.733 \ kW](https://img.qammunity.org/2020/formulas/physics/college/m57swdhuv3iwpcqquwinybwhjj44iw6l9s.png)