Hold a small piece of metal (paper clip?) between solder pads with a tweezer. Heat up tip of a metal fork/spoon (Sharp edge of fork could work better) on a stove. Take it from stove and use it immediately to melt the solder on one pad (seems like there is already enough solder there). I am not sure if it would work but maybe worth trying. Heating entire fork could be hard, so maybe something with smaller metal mass could work better, like a screwdriver, nail clipper etc).

DC motors have high inductance, meaning that the current going over it will resist to change. When you turn off a pair of nmos, current will likely start flowing over the the other pair, from source to drain. Depending on the spec of your nmos, you may consider using diodes in parallel to nmos to carry this current. Obviously these diodes should be reverse biased during normal operation.

Not answering the question but if you studied in English outside of the UK it may be enough. But you need to certify that you studied in English through https://www.ecctis.com/visasandnationality . That is how I did it, and I think it is easier and can be quicker.

One-time programmable (memory)

https://xkcd.com/927/

Paywall

In my first ever programming class textbook was using Allman. Probably for this reason, it is easy for a beginner to match braces. It is a lot loss common industry to my knowledge.

If you are working in a decent workplace, you will receive lots of feedback on your code and what you do. Don’t take it personal and learn from them. Sometimes there are multiple correct answers and yours can be one of them, but each workplace, project and senior colleague has their own concerns and priorities. Sometimes feedback seems to be on a trivial mundane detail, and sometimes it really will be. If you think it is valuable feedback, learn. If you disagree, discuss. Enjoy!

As you said before power on capacitor is discharged. Right after power on capacitor is still discharged, so voltage on capacitor is zero, so reset pin has Vcc. With time capacitor gets charges and voltage across capacitor increases and reset voltage becomes closer and closer to ground, until it is ground. But it is important to consider what happens at power down too. At power down capacitor is charged. If power source becomes high impedance at power down, then reset pin will probably go down to zero in time but may take a bit time depending on what source exactly does. But if power source is connected to zero at power down reset pin will observe minus vcc and slowly go up to 0. If reset pin is sensitive it may be a good idea to protect it with a diode.

Simpsons predicted it once again.

If you knew about the birds and the bees, you would know that this wasn’t random.

Immerse yourself into technology. Become the screen.

I don’t know what it is. It just reminded me of ATMEL8051 and I wanted to share.

I used Atmel8051 in college. It fits nicely on a breadboard and teaches you how to use assembly and make wonders with 512 byte (yes byte) RAM if I remember the number correctly. I think half of that RAM was even reserved.

To be fair 10^(0.000000000000000000001x) is also exponential growth. And if status quo is x=0 and removing entire management means x=10 this means even the max we can get is very little improvement. It can be “exponential” and still not so much.

“Exponentially” is not synonymous to “a lot”. Exponent is a mathematical term and exponential growth requires at least two variables exponentially related to each other. For this to be possibly exponential growth a) progress should be quantifiable (removing management and treating workers well should be quantized somehow) b) performance should be quantifiable and measured at a bunch of progress points (if you have only two measurements it can as well be linear) c) performance should be or can be modeled as a an exponential function of progress in removing management and treating workers well.

I wish we had an active aoe2 community.

Leakage resistance also contributes to dissipation factor and the simple formula omits this, that is why ESR calculated from dissipation factor is larger. As you said, if one is more interested in heat generated, dissipation factor is more important (leakage also dissipates power). If interested in the decoupling and filter performance of the capacitor, ESR is more important. And all these depend on temperature and capacitor bias voltage as well :)

I don’t think this will work well and others already explained why, but thanks for using this community to pitch your idea. We should have more of these discussions here rather than CEO news and tech gossip.

I guess it can happen if you start moving the bottle forward after you start pouring the water.