## BloomLocke367 one year ago more homework check xp

1. BloomLocke367

2. BloomLocke367

@Shalante

3. anonymous

Hard to see. >.<

4. BloomLocke367

sorryyyyyy

5. anonymous

It requires real effort to look at one line.

6. BloomLocke367

I can see it clearly o.o I'm sorryyyyy

7. anonymous

I do not want to check your work, but just do the problem. (Hard to see your work as well)

8. 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

9. anonymous

Only checked the first 2. Lemme see the rest? And do you want me check all the problems?

10. 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

11. anonymous

12. anonymous

Holy sheet. Another worksheet.

13. 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)

14. anonymous

1. is correct.

15. BloomLocke367

yayyy

16. anonymous

2 and 3 is correct

17. BloomLocke367

woooo I am getting better XD

18. BloomLocke367

you deserve like 50 gazillion medals btw XD

19. anonymous

Lol, no need.

20. BloomLocke367

you're awesome omg

21. anonymous

4 and 5 are correct.

22. BloomLocke367

woooo

23. anonymous

Only to you. :p

24. BloomLocke367

hehe thanks xp I don't see why tho

25. anonymous

6 is correct. 100% on first worksheet. Nice!!!

26. anonymous

Hey, I forgot to check sf. 1 and 2 are correct but not the right # of sf.

27. BloomLocke367

wooooooo

28. 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

29. BloomLocke367

1... 50m/s? that drops it to 1 sig fig.. 2...4.0m/s^2?

30. anonymous

I'll check the sf at the end. For 3. You can use only one formula to find the answer. $x _{f}=x _{i}+v _{_{i}}t+\frac{ 1 }{ 2 }at^2$

31. BloomLocke367

I haven't seen that formula before O.O

32. BloomLocke367

We only have 4

33. BloomLocke367

nevermind.. yes I have XD

34. BloomLocke367

lol

35. BloomLocke367

but I still got the right answer, right?

36. anonymous

Same goes for 5. 4 is correct.

37. anonymous

Wait nvm you would still need 2. Your way is faster for 5.

38. BloomLocke367

haha thanks xp

39. BloomLocke367

did I get 3 right?

40. anonymous

You're not stuck on 6 right? You are just taking a break?

41. anonymous

Yes.

42. anonymous

Plug in the values to the equation I gave you and it would still be the same with by just using 1 formula.

43. BloomLocke367

I'm on 8 right now lol

44. anonymous

Nice!

45. BloomLocke367

I got 1.5m for 6 and -0.13m/s^2

46. BloomLocke367

for 7

47. BloomLocke367

well I got the final velocity to be 9.6m/s then I used the formula $$v^2=v_0^2+2a(x-x_0)$$

48. 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.

49. anonymous

50. 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?

51. anonymous

It speeds up by 3.2m/s^2 every second

52. BloomLocke367

OMG I SEE MY MISTAKE

53. BloomLocke367

I didn't square 9.6 lol

54. BloomLocke367

I dunno if I got 8 correct... i got 25m/s after I rounded

55. anonymous

lol

56. anonymous

7 is correct.

57. anonymous

Foe 8, I got the same results.

58. BloomLocke367

oki yay

59. BloomLocke367

I'm starting on 9 now

60. BloomLocke367

I got no for 9.. because the final velocity is 5m/s

61. BloomLocke367

is that right?

62. BloomLocke367
63. anonymous

LMAO

64. anonymous

For 9, I got no because it would be at 25m/s 0.0

65. BloomLocke367

that's what I meant lol

66. BloomLocke367

sorry XD I have issues typing the right numbers for some reason XD

67. anonymous

Lol, no wonder.

68. BloomLocke367

what would I do for 10?

69. anonymous

Hint: one formula.

70. BloomLocke367

hang on.. I may be able to figure it out

71. BloomLocke367

the formula you showed me right..

72. BloomLocke367

?

73. anonymous

yes

74. BloomLocke367

gimme a sec.. everyone is talking

75. BloomLocke367

you can't add t and t^2, can you?

76. BloomLocke367

OMG I KNOW WHAT TO DO.. I CAN USE THE QUADRATIC FORMULA

77. BloomLocke367

wait.. I got both negative answers..but it's time.. @Shalante can you walk me through this?

78. BloomLocke367

wait... nevermind xp I graphed it and got 4.. is it 4 seconds?

79. 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.

80. anonymous

If you are given everything except time in the formula, then it would be a good idea to use that 1 formula.

81. BloomLocke367

ok.. sorry I'm getting stressed

82. BloomLocke367

so which one?

83. BloomLocke367

ohhh do the two separate ones like I have been?

84. anonymous

You got 4 right. Did you do it using 1 formula or 2?

85. BloomLocke367

I graphed the quadratic from the first equation you had me use xp

86. anonymous

Either way works but the one formula is annoying since it has a quadratic.

87. anonymous

Dont do that next time. I messed up also because I hinted you to do it in 1 formula instead of 2.

88. anonymous

vf^2=v^i+2ax then vf=vi+at

89. BloomLocke367

bc I tried the quadratic formula but messed up then I just graphed it XD

90. BloomLocke367

OMG I FOUND MY MISTAKE AND GOT IT WITH THE QUADRATIC FORMULA :D

91. anonymous

Ok next question.

92. BloomLocke367

ok.. idk what to do here

93. anonymous

Are you focusing better right now? Because this question is similar to the ones you done.

94. BloomLocke367

no I'm not tbh

95. BloomLocke367

I GOT IT

96. BloomLocke367

i got 3 seconds

97. BloomLocke367

make that 3.0

98. anonymous

Yea I got 3.0 seconds too.

99. BloomLocke367

okay.. time for the next one.. thank you so much by the way XD

100. BloomLocke367

for 12.. is 0 the initial velocity and 25 the final?

101. anonymous

Yes.

102. BloomLocke367

okay thank you

103. anonymous

No problem!

104. BloomLocke367

oops I forgot x.x

105. BloomLocke367

I got 2083.3m....

106. BloomLocke367

idk if that's right

107. anonymous

Yes, but sig figs makes it 2.1 x 10^3m

108. 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

109. BloomLocke367

okay :)

110. BloomLocke367

LAST ONE.

111. BloomLocke367

then I can move onto precal XD

112. BloomLocke367

omg can I just use a ratio for the last one?

113. BloomLocke367

$\frac{ 53 }{ 120}=\frac{ 85 }{ x }$

114. BloomLocke367

can I do that?

115. anonymous

Yea, wait after, because I am helping someone on a difficult chem question.

116. BloomLocke367

ok

117. anonymous

No because it does not proportional in that kind of ratio.

118. BloomLocke367

what do you mean?

119. 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.

120. BloomLocke367

oh

121. BloomLocke367

nevermind then

122. BloomLocke367

hmmm

123. BloomLocke367

what do I do?

124. anonymous

THINK!!!!

125. BloomLocke367

OH

126. BloomLocke367

wait.. idk

127. BloomLocke367

are you sure it's not proportional?

128. BloomLocke367

why isn't it proportional?

129. 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.

130. BloomLocke367

um.. don't you need time to find acceleration?

131. 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)

132. BloomLocke367

$$\large a=\frac{v-v_0}{t}$$

133. BloomLocke367

right? and we don't have time... wait.. yes we do... v=d/t

134. BloomLocke367

hang on

135. BloomLocke367

I got 2.26 for the time

136. 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?

137. BloomLocke367

I think I got this.. I gotta go..

138. BloomLocke367

ttyl

139. anonymous

Hint: what formula requires distance, initial and final velocity? (Forgot already!?)

140. anonymous

KK

141. anonymous

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!

142. BloomLocke367

V^2=V_0^2+2a(x-x_0) lol