Question

Questions 1d 1e 2f 2l 12b 12e
http://blogs.canby.k12.or.us/uploads/winegarb/apphysicssummer.pdf
This is a random persons answers

Answer 1

So we'll start with umm, 1d, yah?
\[\Large\rm \frac{1}{R_p}=\frac{1}{4.5\times 10^2}+\frac{1}{9.4\times10^2}\]

Answer 2

Let's get a common a denominator.
They already share a factor of 10^2.
So to get a common denominator, the first fraction needs a 9.4 factor,
while the second fraction needs a 4.5 factor.\[\Large\rm \frac{1}{R_p}=\frac{\color{royalblue}{9.4\times}1}{\color{royalblue}{9.4\times}4.5\times 10^2}+\frac{\color{orangered}{4.5\times}1}{\color{orangered}{4.5\times}9.4\times10^2}\]

Answer 3

you can't just reciprocracte everything?

Answer 4

\(\Large\rm 9.4\times4.5\times10^2=\quad 42.3\times10^2=\quad 4.23\times10^3\)

\[\Large\rm \frac{1}{R_p}=\frac{9.4}{4.23\times10^3}+\frac{4.5}{4.23\times10^3}\]The reason we did this is because now we can combine the numerators.\[\Large\rm \frac{1}{R_p}=\frac{9.4+4.5}{4.23\times10^3}\]

Answer 5

Yah maybe that's easier actually :)
Let's see...

Answer 6

Reciprocating is going to be kind of difficult because of the addition between the terms.
I don't think it's going to work out the way you are imagining it will.

Answer 7

See how I condensed the right side down to a single fraction?
We're allowed to flip each side from there.

Answer 8

\[\Large\rm \frac{1}{R_p}=\frac{13.9}{4.23\times10^3}\quad\to\quad R_p=\frac{4.23\times10^3}{13.9}\]

Answer 9

You were hoping we could just do this?\[\Large\rm \frac{1}{R_p}=\frac{1}{4.5\times 10^2}+\frac{1}{9.4\times10^2}\]And flip everything?\[\Large\rm R_p=4.5\times10^2+9.4\times10^2\]It doesn't work out because of the addition sign :(

Answer 10

That one still kinda confusing? :o

Answer 11

oh... Well thanks.. do you have time to help me out with the next one?

Answer 12

\[\Large\rm e=\frac{1.7\times10^3-3.3\times10^2}{1.7\times10^3}\]Honestly, with these powers of 2 and 3, it might just be easier to take them out of scientific notation.\[\Large\rm e=\frac{1700-330}{1700}\]Easier or no?
Do you understand how I changed the numbers?

Answer 13

oh ya

Answer 14

Thanks

Answer 15

2f is the hardest problem...
Can you help me with that. I have no clue waht to do

Answer 16

\[\Large\rm x=x_o+v_ot+\frac{1}{2}at^2\]This one isn't as bad as it seems.
Subtracting x from each side gives us (I'm going to write them in descending order):\[\Large\rm 0=\frac{1}{2}at^2+v_ot+x_o-x\]Let's group the x_o and x together,\[\Large\rm 0=\frac{1}{2}at^2+v_ot+(x_o-x)\]It'll make them easier to work with.
That's not really anything fancy there, I'm just thinking of that as our constant term.
Instead of two separate constant terms.

Lemme add some color so it's a little easier to see what's going on.

Answer 17

\[\Large\rm 0=\color{royalblue}{\frac{1}{2}a}t^2+\color{royalblue}{v_o}t+\color{royalblue}{(x_o-x)}\]We can see that we now have a quadratic in the standard form:\[\Large\rm 0=\color{royalblue}{a}x^2+\color{royalblue}{b}x+\color{royalblue}{c}\]

Answer 18

To solve it, we throw it into the quadratic formula:\[\Large\rm x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

Answer 19

The standard form \(\Large\rm \color{royalblue}{a}\) is not the same \(\Large\rm a\) as in our equation, maybe I should have used a different letter...

Our a coefficient is actually the entire \(\Large\rm \dfrac{1}{2}a\)

Answer 20

Plugging everything in gives us:\[\Large\rm t=\frac{-v_o\pm\sqrt{v_o^2-4\left(\frac{1}{2}a\right)(x_o-x)}}{2\left(\frac{1}{2}a\right)}\]

Answer 21

It's a little tricky, I know.
Getting everything plugged in correctly.

Answer 22

WoW! Thanks so much!

Answer 23

What's the issue with 2b?
Is it a little confusing since they're using two different k's?

Answer 24

2b?

Answer 25

Oh you said 12b, haha my bad :3

Answer 26

its all good :)

Answer 27

2l
\[\Large\rm \frac{1}{f}=\frac{1}{s_o}+\frac{1}{s_i}\]

Answer 28

Subtracting 1/s_o from each side,\[\Large\rm \frac{1}{s_i}=\frac{1}{f}-\frac{1}{s_o}\]I think the answer the answer in the key is a little sloppy.
That's certainly not how I would leave it.

I would first find a common denominator on the right side,\[\Large\rm \frac{1}{s_i}=\frac{s_o}{f\cdot s_o}-\frac{f}{f\cdot s_o}\]\[\Large\rm \frac{1}{s_i}=\frac{s_o-f}{f\cdot s_o}\]Then reciprocate each side,\[\Large\rm s_i=\frac{f\cdot s_o}{s_o-f}\]

Answer 29

This is similar to what I did in the first problem.
So if you're not comfortable getting a common denominator and all that,
then you can reciprocate immediately from this step:\[\Large\rm \frac{1}{s_i}=\frac{1}{f}-\frac{1}{s_o}\]You want to first think of like this:\[\Large\rm \frac{1}{s_i}=\frac{\frac{1}{f}-\frac{1}{s_o}}{1}\]Reciprocating gives:\[\Large\rm \frac{s_i}{1}=\frac{1}{\frac{1}{f}-\frac{1}{s_o}}\]So that's how the key arrived at that solution.

Answer 30

okay that makes sense

Answer 31

\[\Large\rm s_i=\frac{1}{\frac{1}{f}-\frac{1}{s_o}}\]This form leaves unnecessary fractions in the solution, but maybe that's just preference :)

Answer 32

haha okay :)

Answer 33

Thanks so much! I appreciate all of your help. A true israeli supporter you are hahahah :) Helping the oppressed from being oppressed ;)

Answer 34

hehe. where we at? one left?

Answer 35

12 b and 12 e

Answer 36

also does it matter for head to tail the way I draw it if I get the same vector length and direction?