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anonymous

  • 4 years ago

Does this group also help with physics questions?? i know its a form of mathematics and all but am i able to get help if need be? :)

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  1. anonymous
    • 4 years ago
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    i can try

  2. anonymous
    • 4 years ago
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    There's a physics board, and I can try also.

  3. Preetha
    • 4 years ago
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    Post it in the Physics group.

  4. anonymous
    • 4 years ago
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    i did take physics so

  5. anonymous
    • 4 years ago
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    whats the question

  6. anonymous
    • 4 years ago
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    i have tip

  7. anonymous
    • 4 years ago
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    post it on yahoo answers

  8. anonymous
    • 4 years ago
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    after 3 hrs u usually get answers

  9. anonymous
    • 4 years ago
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    okay thank you guys, i dont have a question just yet, i haven't started any work but if i have and im in need of help then i will ask/post it up.

  10. anonymous
    • 4 years ago
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    will do

  11. anonymous
    • 4 years ago
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    if you don't mind post it to this thread, so i can help. i took physics and got an A in it, so i should be able to help, but alas i have a calculus exam to study for, and if you post it in here i can look at it as you have questions.

  12. anonymous
    • 4 years ago
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    okay thank you (: are you any good with the setting out? because i mainly go wrong when it comes to setting out.

  13. anonymous
    • 4 years ago
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    "Setting out"?

  14. anonymous
    • 4 years ago
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    idk about that either? Sorry :( i am guessing that is college level, i was in high for physics , so that course really did not cover that

  15. anonymous
    • 4 years ago
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    no no high school level. umm you know when im solving or doing an equation i must write the formulas and show you i got a curtain answer within the equation. like it as to be in depth.

  16. anonymous
    • 4 years ago
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    If you could literally just type the exact problem, I could solve it. Honestly, and I mean no offense to this, I have no idea what you're saying.

  17. anonymous
    • 4 years ago
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    okay A top sprinter reaches a velocity of 11.13m.s^-1 within 3.15s of start. Assuming that the acceleration is uniform, calculate its magnitude.

  18. TuringTest
    • 4 years ago
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    acceleration is change in velocity over change in time:\[a=\frac{\Delta v}{\Delta t}\]how much does the velocity change in the problem? how long does that take to happen? plug in the numbers

  19. anonymous
    • 4 years ago
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    Yup. v=v_0+at, or v-v_0=at, or (v-v_0)/t=a

  20. anonymous
    • 4 years ago
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    okay, i understand but how do you figure out the magnitude?

  21. anonymous
    • 4 years ago
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    The "magnitude" of something is just the absolute value. For instance, if I was traveling at a velocity of -x, the magnitude would just be x. Now, the sprinter's acceleration is uniform, i.e., constant. So we can use simple division v/t=a. We know his change in velocity, so plug that in. We know the amount of time it took for that change in velocity, so plug that in. Voila, you have the value for acceleration.

  22. anonymous
    • 4 years ago
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    okay so it'll be Vav= s/t where Vav=11.13m.s^-1 t=3.15s and so s=Vav/t s=11.13m.s^-1/ 3.15s therefore s=3.53m.s^-2 is that right?

  23. anonymous
    • 4 years ago
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    As long as you did the division correctly, yeah, that's right. Remember that your "s" is actually acceleration, though.

  24. anonymous
    • 4 years ago
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    is acceleration the same as displacement?? like same concept?

  25. anonymous
    • 4 years ago
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    No. Think about it this way. Movement is displacement. Displacement over time is velocity. Velocity over time is acceleration.

  26. anonymous
    • 4 years ago
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    Therefore, movement over time squared is acceleration.

  27. anonymous
    • 4 years ago
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    ohh okay i get it (: thank you

  28. anonymous
    • 4 years ago
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    i proberly sound really dumb by asking this but how do you find the uniform acceleration?

  29. anonymous
    • 4 years ago
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    Alright, so uniform acceleration (I think) means constant acceleration. That means it doesn't change over time in the system you're given. If there's constant acceleration, it must be the same as the average acceleration. After all, the average of the set (1,1,1,1) is going to be 1, right? Now, your average acceleration is going to be your change in velocity over time. So that means if I started at 5mph, and I went up to 8mph over 10 hours, my change in velocity will 3mph. This was done over 10 hours, so my change in velocity (3mph) over 10 hours will make my acceleration 3/10 or 0.3 repeating miles per hour squared.

  30. anonymous
    • 4 years ago
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    oh okay so say a car has a velocity of 27 m.s and 23s later it increased to 96m.s the uniform acceleration would be 96m.s-27m.s/23s??

  31. anonymous
    • 4 years ago
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    Remember to put parentheses around 96-27 or else your calculator will interpret it badly. But yes. Also remember that this gives you the average acceleration, which is only equal to regular acceleration if the acceleration is uniform or constant.

  32. anonymous
    • 4 years ago
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    okay, ill remember. okay so the equation only "tells" you the regular acceleration if you have uniform acceleration? sorry im just trying to understand it a bit better.

  33. anonymous
    • 4 years ago
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    Yes. Think about it this way: your velocity v is actually a set of numbers telling you your velocity at any given time t. When I have a set of velocities, and I divide them by time, I will get the average velocity, right? Not the velocity at EVERY given point of time, just the average one. It's like if I told you how much money I made over a period time. Let's say I made 10 dollars in 5 days. On average, I made 2 dollars per day if I divide 10/5. But I could've made 7 dollars the first day and 3 dollars the next four. Generally, though, I did make 2 dollars a day on average. But imagine if I told you I made the same amount of money each day, i.e., a constant salary. Then I would have made 2 dollars a day, right? So if I have a constant acceleration, average acceleration is going to be the actual acceleration.

  34. anonymous
    • 4 years ago
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    oh okay i get it now. thank you.

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