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Good (free) digital scientific calculator?
Magus`
The fun has been DOUBLED!Registered User regular
The fun has been DOUBLED!Registered User regular
And yes I know that Windows has one.
I'm having to do some physics homework and I can't find my calculator. Sadly, the Windows one just isn't doing what I want it to do so I'm wondering what other choices I have.
This is due tomorrow so quick answers are a plus.
I'm having to do some physics homework and I can't find my calculator. Sadly, the Windows one just isn't doing what I want it to do so I'm wondering what other choices I have.
This is due tomorrow so quick answers are a plus.
Magus` on
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Posts
That one is made by Microsoft, but it's got a few more features.
Edit: You could try one of these, but I can't vouch for their quality.
I don't believe it - I'm on my THIRD PS3, and my FIRST XBOX360. What the heck?
http://www.7stones.com/Homepage/Publisher/Calc.html
these might work.
im doing physics right now as well. weird.
i think theres also a windows tweakXP calc
what do you need it to do?
That's the one I just linked above. But yes, I should've asked. What specifically do you need?
I don't believe it - I'm on my THIRD PS3, and my FIRST XBOX360. What the heck?
What're you looking to be able to do?
I don't know of resources because most of the time if you've got something complicated enough that you're doing it on a pc, you're using an actual mathematics/statistical language (like R, which is a free download)
However, when I follow what the book tells me, I get the following amount:
9829878.5762342613654567007336687
Now, I'm apparently missing a step.
What the book is telling me to do is the following:
((6.67*10^-11)*(5.98*10^24))/6370^2
It states that will be equal to 9.8m/s^2.
As you can see, it does not.
WHAT AM I MISSING.
Image of the equation: http://teacher.nsrl.rochester.edu/phy_labs/Measure_g/Measure_g04.gif
Steam Profile | Signature art by Alexandra 'Lexxy' Douglass
What're G/M/R in that equation (it's been a while since I did physics)?
Your equation does come out to 9.8*10^6... my guess is some number you're using is in km/sec instead of m/sec or something like that
However, notice that the answer you get is 9829878.5762342613654567007336687?
Assuming that the book is actually telling you to do something like multiply the gravitational constant G by the Earth's radius R (just a guess), I'm willing to bet that you have mismatched units somewhere in there.
By that I mean that you may be using a value in meters when the equation is set up for Kilometers, or something similar. Given that you are 10^6 off, I'm pretty sure that either one of your things is a million times the unit it should be or that the value in the denominator is 1000 times off, which would square to a million.
M = Mass (of the Earth, in this case) (5.98*10^24)
R = Radius from the centre of the Earth (6370^2)
The full equation is basically:
6.67*10^-11 Newtons*Meters^2/Kilograms^2
Times
5.98*10^24Kilograms divided by 6370Kilometres^2
Steam Profile | Signature art by Alexandra 'Lexxy' Douglass
Yep, that's your problem
the radius of earth is 6371.1 KILOMETERS, not meters.. so 6371000 meters (which you need to use since your G number is in meters)
mass is in kg, G is m^3 / (kg * s^2), so you need to use the radius of earth in meters instead of kilometers so the units match up - that's why your answer is off by a factor of 1000^2
Found the problem.
G is in Meters.
R is in Kilometers.
Plug in 6370000 into the equation (Earth's radius in meters), and you should get the expected answer.
Edit: Always a step of ahead of me, Gdiguy.
Should be as simple as plugging a 2 in there, though.
Comes out to something like 2.54 m/s^2. That sound about right-ish?
(Line I'm using: (6.67*10^-11)*((5.98*10^24)/(2*6370000)^2) )
Steam Profile | Signature art by Alexandra 'Lexxy' Douglass
from the line youre using.
I'm working on the next two problems.
The first one asks how much KE is lost when two cars collide (but are still moving).
I basically did:
KE.1 = .5(3)(6)^2 = 54J
KE.2 = .5(2)(4)^2 = 16J
So both cars have a total of 70J KE. However, when they combine to form this equation:
KE.3 = .5(5)(2)^2 = 10J
That's a loss of 60J. Unless I'm doing it wrong, I think I've got that one right.
This next one, I'm kind of lost on:
It says a .5kg hockey puck is at rest. If a net force of 0.8N works on it for a distance of 2M, what will the KE be?
Alright, well I know that work is equal to force times distance. So we have work being listed as 1.6Newton-Metres. That part is easy.
Work is also equal to the change in Kinetic Energy. Getting even closer!
However, this is where I'm stuck. I know that Mechanical Energy is equal to KE + GPE.
Gravity Potential Energy is equal to mass x gravity x height OR weight x height.
As far as I can tell, the height would be zero and thus GPE would be zero.
So what I've figured so far is ME = KE(starting) + GPE (which equals 0).
After this I'm not sure what my next step will be. Any pointers?
Steam Profile | Signature art by Alexandra 'Lexxy' Douglass
That's the free calculator program that my calculus teacher from last semester suggested. It should work for stuff like this just fine.
I'm not really sure how you did the second problem, but that seems like a lot of lost kinetic energy- all of that energy would have to have gone into crushing the cars, so it might be good idea to let us know what your givens were at the start of the problem.
For your third problem, remember that work is equal to change in energy. 1 Joule is 1 Newton-Meter.
Edit: Ok, looks like you already have that part. So, what else are you looking for? You have the initial energy and the change in energy.
A 3kg toy car is moving at 6m/s and collides with a 2kg toy car traveling at a speed of 4m/s.
After the collide, they continue on, locked together, at a speed of 2m/s. I'm assuming I just add the weight of the two toy cars together.
My above equations showed how I handled it.
Edit - Also, I'm assuming the answer for the 3rd problem is 1.6 joules. It already had no KE, and as such 1.6 + 0 is 1.6.
Yay?
Steam Profile | Signature art by Alexandra 'Lexxy' Douglass
This won't effect the change in kinetic energy, that's a scalar quantity.