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The Schwarzschild radius R is given by
Rs=2GM/c^2
G=6.67384e-11 m^3 kg^-1 s^-2 (Newtonian constant of gravitation)
M=The mass of the lasagna
c=299792458 m/s (Speed of light through a vacuum)
The mass of a Stouffers™ Party Size Lasagna is 2.55kg. Let's say we want a lasagna tower that could reach the moon. The tower would have to be 384472.282 km high, or 38447228200 cm. Assuming each lasagna is 10.16 cm high, that would require roughly 3,784,176,004 lasagnas. The mass of our tower would be 9,649,648,810 kg. The Schwarzschild radius would then be 1.4331x10^-14 meters or 0.000014331 nanometers.
I'm tempted to do some napkin math and try to guesstimate how many solar masses that black hole would be.
Ok so let's assume that this galaxy is the size of the Milky Way (100k ly), and let's also eyeball that this black hole is roughly 1/5 of the diameter of its galaxy (huge wild assumptions, I know).
Now we take our black hole at 20,000ly across (lol) and plug this into the Schwarzschild Radius formula r = (2 * G * M)/c^2 and we get 3.2 x 10^17 solar masses (320 quadrillion or 320,000,000,000,000,000). That would make this black hole roughly 550,000 times more massive than the entire Milky Way.
Shit's broke, yo.
The lean angle for a continuous steady-state turn is the angle at which, at a given speed, the force of gravity and centrifugal force are in balance. The formula is arctan(v^2 / rg) where v is velocity, r is the corner radius, and g is the gravitational constant.
https://www.fxsolver.com/blog/2015/06/03/motorcycle-leaning-angle/
Regarding your thought experiment, in my mind it would be an easier experiment to start out riding the circle slowly, and increase your speed a little each circuit. As your speed increases, your lean angle will increase to stay on the circle, and you can keep going until your peg is scraping.
That one can calculate the size (Schwarzschild radius) of a black hole based only on its mass. It is only a black hole if all of the mass of the object is within a sphere of that radius.
Using that, I calculated the Schwarzschild radius of an election. I wanted to try to figure out if an electron is black hole.
The answer is 1.35e-57 meters.
Since the planck lengh is 1.6e-35 meters, I came to the conclusion that it does not matter.
That and the conjecture that the entire universe is a black hole.
This list is extremely useful, thanks!
Check also fxSolver as a candidate for the "Graphing & Visualizing Mathematics" category and as a lighter, free alternative to Wolfram Alpha.
Have a look at fxsolver for the long list of ready-made formula calculators and symbolab for the step-by-step solving functionality.
In reality, there's really no direct competitor to the enormous knowledge base that Wolfram is but there are certain features like the formula database or step-by-step calcs which other tools offer completely for free or can do even better.
There's also a decent list of alternatives here.
> Actually you have to factor in the relative size of each star,
Yes of course, it's at the barycenter. I probably shouldn't have been so general.
The formula is here if anyone wants to see it. Plug in the masses and the distance and solve for r1.
https://en.wikipedia.org/wiki/Barycenter#Two-body_problem
edit: Here a calculator for anyone that wants to solve the problem exactly. https://www.fxsolver.com/solve/
Mechanical engineer here, not electrical. I haven't touched the concepts involved since high school physics, so take this with a grain of salt.
The formula for calculating the magnetic field around a straight length of wire seems to be proportional to the current in the wire, and inversely proportional to the distance from the wire, according to the formula here.
Ignoring one bit I'll get to later, at 700 amps (typical current in transmission lines) and a 40ft distance (typical wooden pole height), the magnetic field would be similar to that measured 7 inches from the cord of a hair dryer (10 amps).
However... magnetic field is a vector with a direction associated with it. The direction of that magnetic field depends on the direction of the current traveling through the wire. Since we're dealing with alternating current, that direction oscillates back and forth 60 times per second. Also of note, we're not dealing with a single conductor. While the current is moving in one direction in a snapshot in time in one wire, there's a parallel wire just a couple feet away carrying the same current in the opposite direction. I would think the magnetic fields approximately cancel out from the point of reference of a car on the ground.
An additional component of this thought exercise is that we can move electrons, but we can't create them. Increasing the number of electrons on an object as far as I know would require transferring them via either conduction or arc flash. The transmission lines are electrically isolated from the ground, so no conduction, and the arc flash distance is going to be a matter of millimeters, not feet. I expect that if the magnetic field of the power lines interacted with the electrons on the metal in the car, it would be negligible amounts of shoving them around that returns to equally spaced electrons as soon as the car leaves the magnetic field.
Alright, here's what I think you can do without having to spend 3 months studying GR :) If you wanted to be 100% accurate you'd have to take into account both effects of gravitational time dilation and time dilation due to velocity (like in the askcience comment I linked). However, since US^2 doesn't simulate general relativity, the velocity reading is rather useless for this purpose.
What I personally would find the best solution is to just use the gravitational time dilation calculator on Wolfram Alpha. What you get from that is not the time dilation of something falling in when it is at distance r, but just the time dilation at distance r i.e. if an observer was stationary at this distance. In the Wolfram calculator you're gonna want to use the mass of the black hole as an input. The distance you need to enter in the Wolfram calculator is the distance to the center of the black hole. You have already got the values of the distance to the event horizon, so you need to add the Schwarzschild radius to that. For a black hole with a mass of 10^10 solar masses, the Schwarzschild radius is about 200 au (you can use this calculator if you want to check). These are some values I get:
| distance from center | time dilation |
|---|---|
| R + 1000 au | 1.10 |
| R + 100 au | 1.72 |
| R + 10 au | 4.54 |
| R + 1 au | 14 |
where R is the Schwarzschild radius.
What you best use for the gravitational force depends on what exactly it is you want to show with this. If you want the physical force that someone feels when falling in the black hole, then this is always zero because when falling you feel weightless.
There's really no direct competitor to the enormous knowledge base that Wolfram is but there are certain features like the formula database or step-by-step calculations which other tools offer completely for free or can do even better.
In short, check fxsolver for the long list of ready-made formula calculators and [symbolab]fxsolver for the step-by-step functionality.
There's also a decent alternatives list here.
I think this is what you were looking for:
https://www.fxsolver.com/solve/share/Bgb3Kjm1JXNHJAbXHXTbSA==/
In order to add a range of values you can use the "generate values" button at the bottom of the screen. Based on the feedback we got here we are working to improve this functionality.
When an airy disc is percieved as a point it is considered sharp. The bigger you make the disc or the closer you view it, the more airy it will become.
This is why dof is based on 8x10 at 2x viewing. Will the disc on a 36mm negative be percieved as a point when magnified up to 8x10 and seen from twice the diagonal away with normal eye sight.
Yes, but the effect would be negligible for anything falling through the atmosphere, unless you needed increased precision for some reason. I dont think (high?)school math problems need that precision.
"Gravity decreases with altitude as one rises above the Earth’s surface because greater altitude means greater distance from the Earth’s centre. All other things being equal, an increase in altitude from sea level to 9,000 metres (30,000 ft) causes a weight decrease of about 0.29%. (An additional factor affecting apparent weight is the decrease in air density at altitude, which lessens an object’s buoyancy. This would increase a person’s apparent weight at an altitude of 9,000 metres by about 0.08%)" - https://www.fxsolver.com/browse/formulas/Gravity+Acceleration+by+Altitude
So there is only a 0.21% decrease in weight at 30k feet, which is negligible. But that also means that if someone does want to be precise, they have to take into account both the changes in gravity along a gradient on the way down, *and* the changes in air pressure along a similar gradient. Being 100% precise is impossible AFAIK.
> https://web.mit.edu/16.unified/www/SPRING/propulsion/notes/node103.html > HERE THIS IS THE EQUATION IM Talking ABOUT
https://www.fxsolver.com/browse/formulas/Ideal+rocket+equation+%28Tsiolkovsky+rocket+equation%29
Δv=V_e *ln * (m_o / m_1 )
Do keep in mind that the Kardashev scale is logarithmic.
In other words, each integer is an order of magnitude greater than the previous one.
A K-2 picking a fight with a K-2.5 is going to end very poorly for the aggressor no matter what.
According to this K-scale power consumption calculator using the equation found in the Wikipedia link above:
a K-2 civ uses 1 x 10^26 Watts
a K-2.5 civ uses 1 x 10^31 Watts
That doesn't sound like much considering one number is 5 larger than the next, but that is actually the number of additional zeroes.
To put this a different way, the entire amount of power a K-2 civilization has access to is quite literally a rounding error in the eyes of a K-2.5 civilization.
I don't think we've ever seen the upper limit on what a death star blast can do, if we knew Alderaan's mass we could work it out with the Gravitational Binding Energy equation (here's a handy calculator for that) but if the Death Star blew up Earth it's output would be at least 2.24*10^(32)J or converted to TNT equivalent 53.58 zettatons (55.58 trillion gigatons). Blowing up planet's ain't no joke.
He's definitely not the most attractive of all men on Tinder. Just someone very attractive combined with high exposure due to his good activity on the platform. If you really wanted to find the most attractive dude, a better methodology would be to present the dudes to every girl in every city, and look at the actual swipe right percentages. Do this test for this guy vs. a hypermasculine dude as PPD describes and I have a feeling the results won't be surprising to anyone.
But even this way wouldn't necessarily select the MOST attractive dude. Tinder is binary, kinda like Rotten Tomato ratings. Contrast that to hypothetically making women actually rate men from 1-10 and aggregating that data, like IMDB's true Bayes formula.
Here is the formula for the elevation angle for ballistic projectile. You can google to find how it's derived.
To solve your question you need to replace u- squared (initial velocity) with the velocity of your projectile as it reaches the height of 3.3m being launched at an angle theta.
I tried to do a very rough estimate by comparing the impactor's kinetic energy to the gravitational binding energy of the planetoid (using this calculator):
- A 5000 kg object traveling at 0.899 c has about 6e20 J of kinetic energy.
- Assuming a uniform density of 1 t/m³ a 500 km diameter object has a binding energy of about 7e23 J. This is about 1000 times the impactor's kinetic energy, so it probably wouldn't blow the planetoid apart completely.
The IMDB movie rankings used to have something like that to make sure movies with few votes didn't outrank those with high numbers of votes and a high ranking.
It's described here, but I'm at work and don't have the time to figure out how it works right now.
https://www.fxsolver.com/browse/formulas/Bayes+estimator+-+Internet+Movie+Database+%28IMDB%29
The ISS weighs ~400,000kg. Accelerating at 1g (before any relativistic effects) would require ~4MN of force. Wikipedia says there are some ion drives that have achieved 5N of thrust for 100kW input energy, so let's say the 4MN would take 20kW per newton of thrust, or roughly 80GW of power for 3.58y x 2 for the round trip.
Assuming we could find some 80GW fusion generator that weighs 0kg, we still have to find enough fuel. 80GW * 7.16 years is ~2E19 Joules. You get roughly 6.5E14 J per kg of hydrogen that you fuse into helium, so we would need about 30,000kg of hydrogen to come along for the ride.
Specific impulse of an ion drive is ~10,000 s (one kg of propellant can generate 9.8N of thrust for 10,000s). The math here is a bit tricky, so I just punched the values into this page (https://www.fxsolver.com/solve/) and came up with approximately 9 billion kg of xenon required to generate this type of thrust.
You can't escape the curse of Tsiolkovsky