9V battery is not able to power 2 sensors. How can I do it?
9V batteries are great if you want to make a smoke detector, but pretty terrible for most other things. Firstly they have a very high internal resistance meaning they can usually only source 100mA or so depending on the make/chemistry. Secondly, they have a very low capacity of maybe a few hundred mAh, which means they don't last very long if you are doing anything other than drawing a few milliamps once in a while (e.g. smoke detector!).

If you then couple that with the fact that you are using a linear regulator to bring it down to 5V, you are basically wasting 44% of the energy in the regulator - all a 7805 does is drop the extra voltage over a variable resistance, burning energy away as heat.

I imagine if you were to measure the terminal voltage of the battery with all those sensors connected (even just the one), you will see it drop considerably. With all of them I imagine it will be down at maybe 4-5V, then because the 7805 has about a 1.5V dropout, the output of the regulator will be maybe 3V or so. Hence everything stops working.

For battery applications, you are far better (if you don't want to go down the LiPo/LiIon route) choosing AA or AAA batteries and getting a switching regulator device to bring that up to 5V. You can get breakout boards which will take 1 or 2 AA or AAA batteries and step up the voltage quite readily. They are also not too tricky to build your own - simple switchers that require maybe 4 or 5 external components are quite common.

Why? Well, AA batteries can have capacities in the couple of 1000mAh range, much higher than a 9V battery, have lower internal resistance, and using a switching regulator instead of a linear regulator will be generally more efficient (not always, but in this case almost certainly).