As of yesterday, Dallas' chief IT official refuses to rule out paying a ransom and says "all options are on the table". He said that there were no signs that banking data had been breached but told customers to "watch for any suspicious activity". I called my bank. They were unaware there was...
I'm not on any company network, just a lone, private computer which is a Mac. I also get email on an iPad. I try to be very careful not to just click on unsolicited links in emails. Only once in a great while do I get a message that looks suspicious. I also try not to forward emails with...
Royal does that much of the time but also can gain direct control through other tools. Just found out that the Dallas Central Appraisal District was hit last year by Royal and they actually paid 170k to the hackers. Operations were impacted for 72 days. What a shame.
Who says there is anything to teach or that it was some mistake by a civil servant? I think Russian professionals can hack into a system by themselves.
Any large scale expansion of our technological civilization into the Solar System will have to make use of the Moon as a waystation, a depot, a source of building materials or even water and oxygen. I don't know if there will be a million people there by 2060 but I'd be surprised if there...
Wired article links gangs to Russian Intelligence services and mentions the 'Royal' gang.
https://www.wired.com/story/conti-trickbot-ransomware-sanctions-uk-us/
The city of Dallas has been the target of a Ransomware Attack by criminal hackers for almost a week now. Some city services are still out. The public is not being told much except they think it is by a group called 'Royal'. Fire, Police, ambulance were all compromised as well as the other city...
I think it's getting better. It seemed to always agree with me even when I gave it bad information as a test. Recently it stood its ground and told me I was wrong.
According to Cambridge english dictionary it should be two words. "Firetruck" may be commonly used but it is not the proper spelling.
https://dictionary.cambridge.org/us/dictionary/english/fire-truck
Me: Is " firetruck" a proper word or is it the two words "fire truck" in proper english...
According to post #4 there is little or no tension in the skin till the rods are tightened so to first order assume all the tension is from the rods. Then take your value for the tuned drumhead tension and multiply that by ##2 \pi## and divide into six parts, one for each rod. That should be the...
As it is being explained to me it would seem to be not unlike a rope turned around a pulley 90 degrees. The tension is the same but the direction is changed. Then to first order, the total tension at the hoop edge translates to a force on each rod which if all the rods are tuned to the same...
Welcome to PF! First a question. In a typical tuned drum, do the tension rods hold all of the tension or does the hoop hold most of it and the rods just add some more?
“we do not believe that any type of laboratory-based scenario is plausible” in explaining the origin of the virus. Instead, the scientists strongly favored a natural origin, arguing that the virus likely spilled from bats into humans, possibly by way of an intermediate animal host."
By origin...
We are all space-time travelling even at the rate of 1 sec/sec because the earth moves. But we can assume the time traveller also moves with the earth as necessary, they are where-ever the earth is whenever they are there just as we are all on the surface of the earth moving with that surface...
But you, the time traveller see all times in sequence. But to the person fixed in time, whenever you observe, the machine has already been there and gone. I think.
That doesn't seem inconsistent to me. People outside could in principle see the machine pass through their time but it would generally be moving at such a rate they probably wouldn't notice it.
At the University of Illinois Physics Dept. (back in the ancient days when I was there) you had two chances to pass the qual. But when you first entered they gave you what was called a 'free shot' at the qual. If you pass, you pass but if you fail it does not count against your two chances. You...
I tested that and it does work. It wrote a Python program. It even documents the program!
ChatGPT:
Here's one way to generate a multiplication table of primes up to 31 in Python:
def is_prime(n):
if n <= 1:
return False
for i in range(2, int(n**0.5) + 1):
if n % i == 0:
return False
return...
I asked it to make a simple multiplication tables of primes up to 31 and it got some of the numbers wrong after the first few. It has a real problem multiplying numbers.
As pointed out by others, this problem seems inconsistent. Where did this problem come from? Homework assignment or did you found it on your own? What is the exact statement of the problem? Thanks.
Remember that ##f(z)## must be of the form $$f(z)=A_zsin(k_z z) + B_zcos(k_z z)$$ because it must be a solution of the separated Helmholtz equation ##\nabla^2_z u_z+ k_z^2 u_z=0##.
However, interestingly, one of our previous solutions above was $$u_z=- \frac{\cos(k_z)}{\sin(k_z)}\sin(k_z z) +...
I think now that won't work because the ##k's## have to be the same for all z terms or it fails the Helmholtz equation because there would be cross terms that do not cancel out so a series solution like
$$u(x,y,z)= \sin(\pi x)\cos(\pi y) \sum_n C_n\cos(k_z z)$$ works but there could be...
Thanks, I missed that, the constant is always 1. In fact this boundary condition implies that while the solution could be
$$u(x,y,z)_n=\sin(\pi x)\cos(\pi y)\cos\left( (n-\frac{1}{2})\pi z \right)$$ for any n it cannot be a linear sum of solutions and meet that boundary condition else we would...
It's not independent of ##z##. I put it in here
$$u(x,y,z)_n= C_n\sin(\pi x)\cos(\pi y) \cos(\frac{n\pi z}{2})$$ but that should be
$$u(x,y,z)_n= C_n\sin(\pi x)\cos(\pi y)\cos\left( (n-\frac{1}{2})\pi z \right)$$
So given the Helmholtz equation $$\nabla^2 u(x,y,z) + k^2u(x,y,z)=0$$ we do the separation of variables $$u=u_x(x)u_y(y)u_z(z)= u_xu_yu_z$$ and ##k^2 = k_x^2 + k_y^2 +k_z^2## giving three separate equations; $$\nabla^2_x u_x+ k_x^2 u_x=0$$ $$\nabla^2_y u_y+ k_y^2 u_y=0$$ $$\nabla^2_z u_z+ k_z^2...