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Good (free) digital scientific calculator?

Magus`Magus` The fun has been DOUBLED!Registered User regular
edited October 2007 in Help / Advice Forum
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.

Magus` on

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    embrikembrik Registered User regular
    edited October 2007
    You could try the MS Power Calculator found here. (Scroll down a little)

    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.

    embrik on
    "Damn you and your Daily Doubles, you brigand!"

    I don't believe it - I'm on my THIRD PS3, and my FIRST XBOX360. What the heck?
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    delphinusdelphinus Registered User regular
    edited October 2007
    http://www.edumedia-sciences.com/a207_l2-scientific-calculator.html

    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?

    delphinus on
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    embrikembrik Registered User regular
    edited October 2007
    delphinus wrote: »
    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?

    embrik on
    "Damn you and your Daily Doubles, you brigand!"

    I don't believe it - I'm on my THIRD PS3, and my FIRST XBOX360. What the heck?
  • Options
    GdiguyGdiguy San Diego, CARegistered User regular
    edited October 2007
    Magus` wrote: »
    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.

    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)

    Gdiguy on
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    Magus`Magus` The fun has been DOUBLED! Registered User regular
    edited October 2007
    Alright, I've been working on this problem. In order to make sure I'm in the right page, I tried to find the acceleration due to gravity (which is 9.8m/s^2).

    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

    Magus` on
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    GdiguyGdiguy San Diego, CARegistered User regular
    edited October 2007
    Magus` wrote: »
    Alright, I've been working on this problem. In order to make sure I'm in the right page, I tried to find the acceleration due to gravity (which is 9.8m/s^2).

    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

    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

    Gdiguy on
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    ChenjesuChenjesu Registered User regular
    edited October 2007
    I ran your calculation as stated through Excel (easy to keep the separate parts straight), and I can confirm your answer to that equation.

    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.

    Chenjesu on
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    Magus`Magus` The fun has been DOUBLED! Registered User regular
    edited October 2007
    G = Universal gravity (6.67*10^-11)
    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

    Magus` on
  • Options
    GdiguyGdiguy San Diego, CARegistered User regular
    edited October 2007
    Magus` wrote: »
    G = Universal gravity (6.67*10^-11)
    M = Mass (of the Earth, in this case) (5.98*10^24)
    R = Radius from the centre of the Earth (6370^2)


    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

    Gdiguy on
  • Options
    ChenjesuChenjesu Registered User regular
    edited October 2007
    Magus` wrote: »
    G = Universal gravity (6.67*10^-11)
    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

    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.

    Chenjesu on
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    Magus`Magus` The fun has been DOUBLED! Registered User regular
    edited October 2007
    Oh doh. I can't believe I missed that. The worst part was I have to figure out the acceleration if it's -2- Radius away.

    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) )

    Magus` on
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    delphinusdelphinus Registered User regular
    edited October 2007
    im getting 2.457
    from the line youre using.

    delphinus on
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    Magus`Magus` The fun has been DOUBLED! Registered User regular
    edited October 2007
    I rounded up.

    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?

    Magus` on
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    TarantioTarantio Registered User regular
    edited October 2007
    http://www.graphcalc.com/

    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.

    Tarantio on
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    Magus`Magus` The fun has been DOUBLED! Registered User regular
    edited October 2007
    For the second problem it's like this:

    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?

    Magus` on
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    TarantioTarantio Registered User regular
    edited October 2007
    Yeah, that all looks good, then. Although for more complicated things, I might be a good idea to make one of the velocities negative, since the two original velocities are in opposite directions.

    This won't effect the change in kinetic energy, that's a scalar quantity.

    Tarantio on
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