BloomLocke367
  • BloomLocke367
more homework check xp
AP Physics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
BloomLocke367
  • BloomLocke367
BloomLocke367
  • BloomLocke367
@Shalante
anonymous
  • anonymous
Hard to see. >.<

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More answers

BloomLocke367
  • BloomLocke367
sorryyyyyy
anonymous
  • anonymous
It requires real effort to look at one line.
BloomLocke367
  • BloomLocke367
I can see it clearly o.o I'm sorryyyyy
anonymous
  • anonymous
I do not want to check your work, but just do the problem. (Hard to see your work as well)
BloomLocke367
  • BloomLocke367
wait... my work is hard to see.. but I can see the problems XD If you get something different I'll just ask about it lol
anonymous
  • anonymous
Only checked the first 2. Lemme see the rest? And do you want me check all the problems?
BloomLocke367
  • BloomLocke367
I would like it.. but you don't have to lol... I think I'm doing a better job than last time with this stuff
anonymous
  • anonymous
Glad to hear that.
anonymous
  • anonymous
Holy sheet. Another worksheet.
BloomLocke367
  • BloomLocke367
yea.. 19 problems for one night... plus a worksheet from last week that I haven't continued on yet lol (that's the one you helped with last time lol)
anonymous
  • anonymous
1. is correct.
BloomLocke367
  • BloomLocke367
yayyy
anonymous
  • anonymous
2 and 3 is correct
BloomLocke367
  • BloomLocke367
woooo I am getting better XD
BloomLocke367
  • BloomLocke367
you deserve like 50 gazillion medals btw XD
anonymous
  • anonymous
Lol, no need.
BloomLocke367
  • BloomLocke367
you're awesome omg
anonymous
  • anonymous
4 and 5 are correct.
BloomLocke367
  • BloomLocke367
woooo
anonymous
  • anonymous
Only to you. :p
BloomLocke367
  • BloomLocke367
hehe thanks xp I don't see why tho
anonymous
  • anonymous
6 is correct. 100% on first worksheet. Nice!!!
anonymous
  • anonymous
Hey, I forgot to check sf. 1 and 2 are correct but not the right # of sf.
BloomLocke367
  • BloomLocke367
wooooooo
BloomLocke367
  • BloomLocke367
and he doesn't really care about sig figs right now but thanks c: I try getting them anyways for practice. I'll look and fix it
BloomLocke367
  • BloomLocke367
1... 50m/s? that drops it to 1 sig fig.. 2...4.0m/s^2?
anonymous
  • anonymous
I'll check the sf at the end. For 3. You can use only one formula to find the answer. \[x _{f}=x _{i}+v _{_{i}}t+\frac{ 1 }{ 2 }at^2\]
BloomLocke367
  • BloomLocke367
I haven't seen that formula before O.O
BloomLocke367
  • BloomLocke367
We only have 4
BloomLocke367
  • BloomLocke367
nevermind.. yes I have XD
BloomLocke367
  • BloomLocke367
lol
BloomLocke367
  • BloomLocke367
but I still got the right answer, right?
anonymous
  • anonymous
Same goes for 5. 4 is correct.
anonymous
  • anonymous
Wait nvm you would still need 2. Your way is faster for 5.
BloomLocke367
  • BloomLocke367
haha thanks xp
BloomLocke367
  • BloomLocke367
did I get 3 right?
anonymous
  • anonymous
You're not stuck on 6 right? You are just taking a break?
anonymous
  • anonymous
Yes.
anonymous
  • anonymous
Plug in the values to the equation I gave you and it would still be the same with by just using 1 formula.
BloomLocke367
  • BloomLocke367
I'm on 8 right now lol
anonymous
  • anonymous
Nice!
BloomLocke367
  • BloomLocke367
I got 1.5m for 6 and -0.13m/s^2
BloomLocke367
  • BloomLocke367
for 7
BloomLocke367
  • BloomLocke367
well I got the final velocity to be 9.6m/s then I used the formula \(v^2=v_0^2+2a(x-x_0)\)
anonymous
  • anonymous
9.6^2=92.16m 92.16m=0+2(3.2m/s^2)(d) 92.16m/6.4m/s^2=14.4m or 14m when rounded to sf.
anonymous
  • anonymous
You can tell your answer is wrong by reading the problem and thinking like this :
anonymous
  • anonymous
Hmm, if a car accelerate for 3.2m/s^2 for 3 sec straight, wouldn't the distance be more than like around 5m?
anonymous
  • anonymous
It speeds up by 3.2m/s^2 every second
BloomLocke367
  • BloomLocke367
OMG I SEE MY MISTAKE
BloomLocke367
  • BloomLocke367
I didn't square 9.6 lol
BloomLocke367
  • BloomLocke367
I dunno if I got 8 correct... i got 25m/s after I rounded
anonymous
  • anonymous
lol
anonymous
  • anonymous
7 is correct.
anonymous
  • anonymous
Foe 8, I got the same results.
BloomLocke367
  • BloomLocke367
oki yay
BloomLocke367
  • BloomLocke367
I'm starting on 9 now
BloomLocke367
  • BloomLocke367
I got no for 9.. because the final velocity is 5m/s
BloomLocke367
  • BloomLocke367
is that right?
BloomLocke367
  • BloomLocke367
OMG http://prntscr.com/8e69uy
anonymous
  • anonymous
LMAO
anonymous
  • anonymous
For 9, I got no because it would be at 25m/s 0.0
BloomLocke367
  • BloomLocke367
that's what I meant lol
BloomLocke367
  • BloomLocke367
sorry XD I have issues typing the right numbers for some reason XD
anonymous
  • anonymous
Lol, no wonder.
BloomLocke367
  • BloomLocke367
what would I do for 10?
anonymous
  • anonymous
Hint: one formula.
BloomLocke367
  • BloomLocke367
hang on.. I may be able to figure it out
BloomLocke367
  • BloomLocke367
the formula you showed me right..
BloomLocke367
  • BloomLocke367
?
anonymous
  • anonymous
yes
BloomLocke367
  • BloomLocke367
gimme a sec.. everyone is talking
BloomLocke367
  • BloomLocke367
you can't add t and t^2, can you?
BloomLocke367
  • BloomLocke367
OMG I KNOW WHAT TO DO.. I CAN USE THE QUADRATIC FORMULA
BloomLocke367
  • BloomLocke367
wait.. I got both negative answers..but it's time.. @Shalante can you walk me through this?
BloomLocke367
  • BloomLocke367
wait... nevermind xp I graphed it and got 4.. is it 4 seconds?
anonymous
  • anonymous
You would need quadratic, which is a pain. So the one formula I told you to use was bad. vf^2=v^i+2ax then vf=vi+at is better.
anonymous
  • anonymous
If you are given everything except time in the formula, then it would be a good idea to use that 1 formula.
BloomLocke367
  • BloomLocke367
ok.. sorry I'm getting stressed
BloomLocke367
  • BloomLocke367
so which one?
BloomLocke367
  • BloomLocke367
ohhh do the two separate ones like I have been?
anonymous
  • anonymous
You got 4 right. Did you do it using 1 formula or 2?
BloomLocke367
  • BloomLocke367
I graphed the quadratic from the first equation you had me use xp
anonymous
  • anonymous
Either way works but the one formula is annoying since it has a quadratic.
anonymous
  • anonymous
Dont do that next time. I messed up also because I hinted you to do it in 1 formula instead of 2.
anonymous
  • anonymous
vf^2=v^i+2ax then vf=vi+at
BloomLocke367
  • BloomLocke367
bc I tried the quadratic formula but messed up then I just graphed it XD
BloomLocke367
  • BloomLocke367
OMG I FOUND MY MISTAKE AND GOT IT WITH THE QUADRATIC FORMULA :D
anonymous
  • anonymous
Ok next question.
BloomLocke367
  • BloomLocke367
ok.. idk what to do here
anonymous
  • anonymous
Are you focusing better right now? Because this question is similar to the ones you done.
BloomLocke367
  • BloomLocke367
no I'm not tbh
BloomLocke367
  • BloomLocke367
I GOT IT
BloomLocke367
  • BloomLocke367
i got 3 seconds
BloomLocke367
  • BloomLocke367
make that 3.0
anonymous
  • anonymous
Yea I got 3.0 seconds too.
BloomLocke367
  • BloomLocke367
okay.. time for the next one.. thank you so much by the way XD
BloomLocke367
  • BloomLocke367
for 12.. is 0 the initial velocity and 25 the final?
anonymous
  • anonymous
Yes.
BloomLocke367
  • BloomLocke367
okay thank you
anonymous
  • anonymous
No problem!
BloomLocke367
  • BloomLocke367
oops I forgot x.x
BloomLocke367
  • BloomLocke367
I got 2083.3m....
BloomLocke367
  • BloomLocke367
idk if that's right
anonymous
  • anonymous
Yes, but sig figs makes it 2.1 x 10^3m
anonymous
  • anonymous
That means the minimum distance is 2083.3m in order for the train to reach a velocity of 25 m/s. It makes since because the train is accelerating at 0.15 m/s^2
BloomLocke367
  • BloomLocke367
okay :)
BloomLocke367
  • BloomLocke367
LAST ONE.
BloomLocke367
  • BloomLocke367
then I can move onto precal XD
BloomLocke367
  • BloomLocke367
omg can I just use a ratio for the last one?
BloomLocke367
  • BloomLocke367
\[\frac{ 53 }{ 120}=\frac{ 85 }{ x }\]
BloomLocke367
  • BloomLocke367
can I do that?
anonymous
  • anonymous
Yea, wait after, because I am helping someone on a difficult chem question.
BloomLocke367
  • BloomLocke367
ok
anonymous
  • anonymous
No because it does not proportional in that kind of ratio.
BloomLocke367
  • BloomLocke367
what do you mean?
anonymous
  • anonymous
speed 1/distance 1 is not proportional to speed 2/distance 2 Fraction is called a ratio. Meant to write it is not proportional* in the previous post.
BloomLocke367
  • BloomLocke367
oh
BloomLocke367
  • BloomLocke367
nevermind then
BloomLocke367
  • BloomLocke367
hmmm
BloomLocke367
  • BloomLocke367
what do I do?
anonymous
  • anonymous
THINK!!!!
BloomLocke367
  • BloomLocke367
OH
BloomLocke367
  • BloomLocke367
wait.. idk
BloomLocke367
  • BloomLocke367
are you sure it's not proportional?
BloomLocke367
  • BloomLocke367
why isn't it proportional?
anonymous
  • anonymous
Ok, a drag racer can reach a speed of 53 m/s can also mean that when it starts from rest the drag racer can reach a speed of 53 m/s So initial velocity is 0 m/s. Final velocity would be 53 m/s. Distance would be 120m Solve for acceleration using the formula. (Not telling you which formula) But what about 85 m/s. It needs to start from rest in order for the drag racer to move right? Initial velocity is zero, final velocity is 85, acceleration is the one you solved earlier, and distance is what you need to know for the answer.
BloomLocke367
  • BloomLocke367
um.. don't you need time to find acceleration?
anonymous
  • anonymous
We would use the same acceleration since it is the same drag racer and it told us he had a max acceleration (his highest is 53m/s at 120m)
BloomLocke367
  • BloomLocke367
\(\large a=\frac{v-v_0}{t}\)
BloomLocke367
  • BloomLocke367
right? and we don't have time... wait.. yes we do... v=d/t
BloomLocke367
  • BloomLocke367
hang on
BloomLocke367
  • BloomLocke367
I got 2.26 for the time
anonymous
  • anonymous
Why would use that formula when they dont give you the 2 units (time and acceleration) How are you going to solve the unknown variable?
BloomLocke367
  • BloomLocke367
I think I got this.. I gotta go..
BloomLocke367
  • BloomLocke367
ttyl
anonymous
  • anonymous
Hint: what formula requires distance, initial and final velocity? (Forgot already!?)
anonymous
  • anonymous
KK
Jhannybean
  • Jhannybean
Have you made a flashcard with all the kinematic equations on it? If not, I suggest you do, and then either post it on your wall, look at it every day and then see how many you have memorized. Carry the flashcard with you to school as well, and review it every chance you get. The faster you can recall the formulas and how they are applied, the quicker you will finish your test!
BloomLocke367
  • BloomLocke367
V^2=V_0^2+2a(x-x_0) lol

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