Question

@jim_thompson5910

You'll need this to see things

http://cyh.leeschools.net/UserContent/Documents/AP%20Physics1%20SumAssign%2014-15.pdf

Answer 1

I did the first 8, thought you could check them

Answer 2

ok go ahead and post what you got

Answer 3

And by the first 8 I mean 1a-h

Answer 4

a-h = ah

lol sry, random...anyways, go ahead

Answer 5

Okay the first one is obviously \[v=\frac{ x }{ t }\]

Answer 6

yep

Answer 7

1b I got \[a=\frac{ v ^2-v _{0}^{2} }{ 2(x-x _{0}) }\]

Answer 8

good

Answer 9

Yay cool

Answer 10

1c I got\[x=\sqrt{2k(E _{el)}}\]

Answer 11

that's incorrect

Answer 12

Darn alright

Answer 13

you aren't multiplying both sides by k

Answer 14

Oh right so would it be \[x=\frac{ \sqrt{2(E _{el})} }{ k }\]

Answer 15

much better

Answer 16

actually, the square root applies to the k as well, it should be

\[\Large x = \sqrt{ \frac{2E_{el}}{k} }\]

Answer 17

Yeah I know

Answer 18

because you divide both sides by k

then you apply the square root

Answer 19

I just didn't write it like that wasnt sure how to lol

Answer 20

oh then be careful because

\[\frac{ \sqrt{2(E _{el})} }{ k } \neq \sqrt{ \frac{2E_{el}}{k} }\]

Answer 21

oh I gotcha

Answer 22

1d I got (I think this ones wrong) \[g=(\frac{ T }{ l2\Pi})^2\]

Answer 23

That denominator says l2pi it's just not very noticable

Answer 24

anything you're curious on how to write, you can ask me

or you can right-click on the math equation and click "show math as ---> Tex Commands"

Answer 25

Is 1d right?

Answer 26

one sec

Answer 27

Wait I see a flaw in it myself lol

Answer 28

unfortunately it's incorrect

Answer 29

I think it's \[g=(\frac{ T }{ 2\pi })^2*\frac{ 1 }{ l }\]

Answer 30

This is what I get

\[\Large T = 2\pi\sqrt{\frac{L}{g}}\]
\[\Large \frac{T}{2\pi} = \sqrt{\frac{L}{g}}\]
\[\Large \left(\frac{T}{2\pi}\right)^2 = \frac{L}{g}\]
\[\Large \left(\frac{2\pi}{T}\right)^2 = \frac{g}{L}\]
\[\Large L\left(\frac{2\pi}{T}\right)^2 = g\]
\[\Large g = L\left(\frac{2\pi}{T}\right)^2\]

Answer 31

Notice how I flipped both fractions to get \(\Large \frac{g}{L}\)

Answer 32

Yeah why though

Answer 33

because I then multiplied both sides by L to isolate g

Answer 34

to isolate g, it's easier to do so when its in the numerator (not denominator)

Answer 35

I thought when you have (T/2pi)^2=l/g

That your could multiply each side by 1/l because l/g*1/l= l/gl and the l's cancel?

Answer 36

or you can do it like this

\[\Large T = 2\pi\sqrt{\frac{L}{g}}\]
\[\Large \frac{T}{2\pi} = \sqrt{\frac{L}{g}}\]
\[\Large \left(\frac{T}{2\pi}\right)^2 = \frac{L}{g}\]
\[\Large \frac{T^2}{(2\pi)^2} = \frac{L}{g}\]
\[\Large \frac{T^2}{4\pi^2} = \frac{L}{g}\]
\[\Large T^2*g = 4\pi^2*L\]
\[\Large g = \frac{4\pi^2*L}{T^2}\]

Answer 37

But my way doesn't work?

Answer 38

if you did that, you'd have

\[\Large \frac{1}{L}*\left(\frac{T}{2\pi}\right)^2 = \frac{1}{g}\]

Answer 39

so you'd be on your way isolating g, but not quite done yet

Answer 40

Ohhh right

Answer 41

to get g all by itself, you'd just flip everything

Answer 42

to go from

\[\Large \frac{1}{L}*\left(\frac{T}{2\pi}\right)^2 = \frac{1}{g}\]

to

\[\Large \frac{L}{1}*\left(\frac{2\pi}{T}\right)^2 = g\]

Answer 43

Yeah I get it now

Answer 44

ok great

Answer 45

1e I got \[r=\sqrt{m _{1}m _{2}(\frac{ G }{ F _{g}})}\]

Answer 46

good

Answer 47

1f I got \[v=\sqrt{\frac{ 2mgh }{ m }}\]

Answer 48

I gues the m on top would cancel right?

Answer 49

correct, giving you

\[\Large v = \sqrt{2gh}\]

Answer 50

Oh I thought just the top one cancels because all of that is over m?

Or is that if it was like 2m+g+h/m?

Answer 51

it cancels with the m in the bottom as well

they divide to 1

\[\Large \frac{2mgh}{m} = \frac{2\cancel{m}gh}{\cancel{m}} = 2gh\]

Answer 52

Okay I thought so

Answer 53

I'm not sure what you mean by 2m+g+h/m though

Answer 54

Like then it would be 2+g+h/m? Because that 2m+g+h/m is the same as

2m/m+g/m+h/m

Answer 55

oh i see, yeah if we had that then m would stick around somewhat

but in this case we're multiplying and not adding

Answer 56

(2m+g+h)/m would turn into 2m/m+g/m+h/m = 2 + (g/m) + (h/m)

but that's a bit off tangent

Answer 57

1g I got \[t=\sqrt{\frac{ 2(x-x _{0}-v _{0}) }{ a }}\]

Answer 58

correct

Answer 59

And 1h I got \[v=\frac{ r(L) } {m }\]

Answer 60

that's incorrect

Answer 61

you divide both sides by mr to isolate v

Answer 62

Okay yeah I wasn't sure if that's what I had to do

Answer 63

I skipped part 2 because I forgot about it because Chem was awhile ago

But I did Part 3 and the first part of part 4 and you can check that, I dont think im wrong on any of it, but just wanna double check

Answer 64

part isn't so bad really

you are just going from one unit to another and you take advantage of cancellations

for example

to convert 25 km to meters, you multiply by (1000 m)/(1 km)

you'll notice the km unit cancel leaving you with meters

Answer 65

part 2*