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anonymous

  • one year ago

http://prntscr.com/8i2msx

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  1. anonymous
    • one year ago
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    @dan815

  2. anonymous
    • one year ago
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    Anyone there?

  3. dan815
    • one year ago
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    okay so to a Cathode in the CRT, would you normally connect DC current or AC current

  4. anonymous
    • one year ago
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    ok?

  5. dan815
    • one year ago
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    thats a question lol

  6. anonymous
    • one year ago
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    well.... http://www.filedropper.com/physicspapermanekji

  7. dan815
    • one year ago
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    whats that

  8. anonymous
    • one year ago
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    I need help with the entire paper..

  9. dan815
    • one year ago
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    i cant see that

  10. dan815
    • one year ago
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    okay so tell me

  11. anonymous
    • one year ago
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    You'll Have To Download It.

  12. anonymous
    • one year ago
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    http://prntscr.com/8i2tn7 there

  13. dan815
    • one year ago
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    okay look at question 1a

  14. dan815
    • one year ago
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    what happens when current is passed into the chatode

  15. anonymous
    • one year ago
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    idk

  16. dan815
    • one year ago
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    think about it

  17. dan815
    • one year ago
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    what is a cathode

  18. anonymous
    • one year ago
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    negative terminal

  19. dan815
    • one year ago
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    okay so when you complete the circuit

  20. dan815
    • one year ago
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    what will happen between the cathode and the anode

  21. anonymous
    • one year ago
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    what u mean?

  22. dan815
    • one year ago
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    an electron is ejected from the cathode and goes to the anode

  23. anonymous
    • one year ago
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    electron beam?

  24. dan815
    • one year ago
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    yes..

  25. dan815
    • one year ago
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    but you need a high enough voltage to displace an electron from the cathode

  26. dan815
    • one year ago
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    so when you are alternating the current during the lower voltages it will not displace an electron, this way you can control the time betweeen when your electrons are ejected, the frequency of your alternating current give you control on the frequency of ejection timings

  27. dan815
    • one year ago
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    an alternating current is a result of an alternating voltage and vice vera

  28. anonymous
    • one year ago
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    question 1 b?

  29. anonymous
    • one year ago
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    beacuse high resistance right?

  30. dan815
    • one year ago
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    what do you think is the answer to b

  31. dan815
    • one year ago
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    high resistance where

  32. anonymous
    • one year ago
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    High resistance in the wire?

  33. dan815
    • one year ago
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    wire for what

  34. dan815
    • one year ago
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    You're going to have to be more clear if u want my help :)

  35. dan815
    • one year ago
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    I am not going to spoon feed you answers

  36. anonymous
    • one year ago
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    Thick wires are used for lighting purposes because current is less?

  37. anonymous
    • one year ago
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    higher the resistance lesser the current right?

  38. dan815
    • one year ago
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    the question is actually saying thick wires are used for power circuits rather than lighting purpose

  39. dan815
    • one year ago
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    okay lets think about this a little, why does a lightbulb light up

  40. dan815
    • one year ago
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    what is that fillament in the lightbulb really

  41. dan815
    • one year ago
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    |dw:1442671036320:dw|

  42. anonymous
    • one year ago
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    i'm getting confused :(

  43. dan815
    • one year ago
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    the fillament in the lightbulb is basically just a resistor

  44. dan815
    • one year ago
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    its a very good resistor

  45. dan815
    • one year ago
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    what happens in a lightbulb is that you want lots of current to pass through the fillament, you want enough current to pass through that resistor to heat it up enough to make it emit enough photos, ie light

  46. dan815
    • one year ago
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    so you dont want needless current being lost to the wires

  47. dan815
    • one year ago
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    if u have thick wires your circult looks like this

  48. dan815
    • one year ago
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    |dw:1442671207500:dw|

  49. dan815
    • one year ago
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    your current will be the sum of the voltage/sum of all resistances

  50. dan815
    • one year ago
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    you just lowered your whole current needlessly

  51. dan815
    • one year ago
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    and there is less current through the fillament taking longer for it to heat up

  52. dan815
    • one year ago
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    now when it comes to power circuits

  53. dan815
    • one year ago
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    you dont want too much current going through resistors needlessly using up power in your circuit

  54. dan815
    • one year ago
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    also wearing down the life time of your circuit

  55. dan815
    • one year ago
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    so you get thicker wires to take the load off the components in power circuits

  56. anonymous
    • one year ago
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    got it... q1c potential energy is mgh right?

  57. dan815
    • one year ago
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    yes

  58. dan815
    • one year ago
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    make sure your units make sense

  59. anonymous
    • one year ago
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    i need to convert 200 gm to kg right?

  60. dan815
    • one year ago
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    yes

  61. anonymous
    • one year ago
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    but they haven't given the h...

  62. dan815
    • one year ago
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    you know the acceleration due to earths gravity is -9.8m/s^2

  63. anonymous
    • one year ago
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    yes..

  64. dan815
    • one year ago
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    this means you are losing a speed of 9.8 m/s every second

  65. dan815
    • one year ago
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    i will make that 10m/s per second to make it simpler for now, u calcuate the right answer

  66. dan815
    • one year ago
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    okay so suppose your accelerationg is -10m/s/s means u lose 10m/s of speed every second, you began with 20m/s so in 1 second you go from 20 m/s to 10m/s in another second you fo from 10 m/s to 0 m/s in 2 seconds your veloicty is 0m/s so your ball has come to a momentary stop up in the air, before it begins to head down

  67. dan815
    • one year ago
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    any questions so far?

  68. anonymous
    • one year ago
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    PE = mgh i can't see the value of h

  69. dan815
    • one year ago
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    you dont have to use mgh

  70. dan815
    • one year ago
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    you can use normal physics

  71. anonymous
    • one year ago
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    is the ans 4J?

  72. dan815
    • one year ago
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    okay if u wanna use mgh

  73. dan815
    • one year ago
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    you can use kinetic energy and potential energy

  74. dan815
    • one year ago
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    at the bottom there is only kinetic energy 1/2 mv^2 at the highest point there is only potential enrgy mgh

  75. dan815
    • one year ago
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    1/2mv^2=mgh solve for h

  76. anonymous
    • one year ago
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    PE = KE=Mv^2/2?

  77. dan815
    • one year ago
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    okay, but i think you should really learn to solve it the other way too, energy theorem stuff comes after being able to solve the basic mechanics way

  78. dan815
    • one year ago
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    both are very important

  79. anonymous
    • one year ago
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    energy theorem?

  80. anonymous
    • one year ago
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    so m is 0.2 right?

  81. dan815
    • one year ago
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    yeah well you see from the equation the mass irrelevant

  82. dan815
    • one year ago
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    and it make sense, right if you were to drop any mass at some height, it will fall down with the same velocity

  83. dan815
    • one year ago
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    theoretically, but in reality, air resistance wont really allow that

  84. dan815
    • one year ago
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    anyways so 1/2mv^2=mgh mass cancels on both sides

  85. anonymous
    • one year ago
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    2000?

  86. dan815
    • one year ago
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    wait lemme see

  87. dan815
    • one year ago
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    oh thats wrong

  88. dan815
    • one year ago
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    |dw:1442672220046:dw|

  89. anonymous
    • one year ago
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    wait its 20m

  90. dan815
    • one year ago
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    yes

  91. dan815
    • one year ago
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    approximately remember we are using 10 m/s^2 for gravity instead of 9.8m/s^2

  92. dan815
    • one year ago
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    if u have a calculator use the right one for your answer

  93. anonymous
    • one year ago
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    PE is 40J?

  94. dan815
    • one year ago
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    0.2kg*10m/s^2*20m=40j

  95. dan815
    • one year ago
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    yes

  96. anonymous
    • one year ago
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    q1d?

  97. dan815
    • one year ago
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    okay so lets say some unsuspecting person happened to be touching the contact while he was pluggig it in

  98. dan815
    • one year ago
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    more current always wants to pass through the biggest voltage drop

  99. dan815
    • one year ago
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    if the ground is not in before your connection is made the current wants to pass through the person and u get the brunt of the current going through you

  100. dan815
    • one year ago
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    but if the prong is longer then that is always going to be connected inside before the other 2 go in

  101. anonymous
    • one year ago
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    It's q related to sound

  102. dan815
    • one year ago
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    oh that one

  103. anonymous
    • one year ago
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    like wavelength and all

  104. dan815
    • one year ago
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    okay well they gave u the distance and time for that so v=d/t

  105. anonymous
    • one year ago
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    3/12?

  106. dan815
    • one year ago
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    yes

  107. anonymous
    • one year ago
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    0.25 ms^-1?

  108. dan815
    • one year ago
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    1/4 m/s if u reduce

  109. anonymous
    • one year ago
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    q2a?

  110. anonymous
    • one year ago
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    yeah i reduced it to 1/4 then got 0.25 as d ans

  111. dan815
    • one year ago
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    ok u can do the rest, just work on all of them and post your solutions

  112. dan815
    • one year ago
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    i have to do my assignment

  113. anonymous
    • one year ago
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    i'll try but....can u explain q2a to me?

  114. anonymous
    • one year ago
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    wait its 2 m right?

  115. anonymous
    • one year ago
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    moment - force times distance

  116. anonymous
    • one year ago
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    =*

  117. anonymous
    • one year ago
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    0.44

  118. anonymous
    • one year ago
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    @dan815

  119. dan815
    • one year ago
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    whaat -.-

  120. dan815
    • one year ago
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    i got my own hw too

  121. anonymous
    • one year ago
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    the ans to q2a is 0.447 m right?

  122. dan815
    • one year ago
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    no

  123. anonymous
    • one year ago
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    whaat-.-

  124. dan815
    • one year ago
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    moment = force * distance 960Nm=430N * x

  125. anonymous
    • one year ago
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    wait moment = force*distance so 960 = 430 times d 2.32?

  126. dan815
    • one year ago
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    yes

  127. anonymous
    • one year ago
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    ok thanks ..

  128. dan815
    • one year ago
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    lol ..

  129. anonymous
    • one year ago
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    i'll tag u if i need help

  130. dan815
    • one year ago
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    -.-

  131. dan815
    • one year ago
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    i wont come

  132. anonymous
    • one year ago
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    :(

  133. dan815
    • one year ago
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    hahahah

  134. dan815
    • one year ago
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    just use youtube or google man

  135. dan815
    • one year ago
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    thats how we all learnt when we got stuck

  136. anonymous
    • one year ago
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    thanks..

  137. anonymous
    • one year ago
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    @hartnn

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