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anonymous

  • one year ago

Cleo and Clare are looking from their balcony to a swimming pool below that is located 15 m horizontally from the bottom of their building. They estimate the balcony is 45 m high and wonder how fast they would have to jump horizontally to succeed in reaching the pool. What calculations would you show to help them determine the answer? Evaluate the practicality of their being able to succeed at jumping into the pool.

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  1. anonymous
    • one year ago
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    @iGreen

  2. anonymous
    • one year ago
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    @Michele_Laino

  3. anonymous
    • one year ago
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    @radar

  4. anonymous
    • one year ago
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    In three scenarios a truck is towing a boat on a trailer. Using 3 – 4 complete sentences indicate which scenario(s) (if any) is the force of the truck on the trailer greater than the force of the trailer on the truck: Scenario A: masstruck = 1000 kg, massboat and trailer = 500 kg; velocity = 20 m/s Scenario B: masstruck = 800 kg, massboat and trailer = 800 kg; velocity = 20 m/s Scenario C: masstruck = 500 kg, massboat and trailer = 1000 kg; velocity = 15 m/s

  5. Michele_Laino
    • one year ago
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    hello!

  6. Michele_Laino
    • one year ago
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    have you got any ideas?

  7. anonymous
    • one year ago
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    Not at all, i got some some earlier but theres a few that I'm lost on.

  8. Michele_Laino
    • one year ago
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    I think that in all of your three cases, both forces have the same magnitude, since we have to apply the third law of Newton

  9. Michele_Laino
    • one year ago
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    furthermore, since the velocity is constant, then we can consider the system truck+boat, an inertial system

  10. Michele_Laino
    • one year ago
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    and, as you know, the laws of Physics are the same into all inertial systems

  11. anonymous
    • one year ago
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    so none of them are greater than the other? correct?

  12. Michele_Laino
    • one year ago
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    yes! correct, since the velocity of the system truck+boat, is constant in all of your three cases

  13. anonymous
    • one year ago
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    how could i try and explain that in a few sentences?

  14. Michele_Laino
    • one year ago
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    here is my sentence: "since the velocity of the system truck+boat" is constant, as vector, then the same system "truck+boat" is an inertial system. Now within an inertial system, all law of Physics, have the same shape, as formula, so we can apply the third law of Newton, and we can conclude that the force applied by truck on boat have the same direction and magnitude of the force applied by the boat on truck. Furthermore those forces have opposite orientation each other"

  15. anonymous
    • one year ago
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    what about the first question?

  16. Michele_Laino
    • one year ago
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    we have to consider these two equations: \[\left\{ \begin{gathered} h = \frac{1}{2}g{t^2} \hfill \\ \hfill \\ d = vt \hfill \\ \end{gathered} \right.\] where d=15 meters and h= 45 meters

  17. Michele_Laino
    • one year ago
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    now, I solve the second equation for t, and I get: \[t = \frac{d}{v}\]

  18. Michele_Laino
    • one year ago
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    Now I substitute into the first equation, and I get: \[45 = \frac{1}{2}g{\left( {\frac{{15}}{v}} \right)^2}\] please solve that equation for velocity v, here g=9.81 m/sec^2

  19. anonymous
    • one year ago
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    i dont understand you math I'm confused

  20. anonymous
    • one year ago
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    @Michele_Laino

  21. Michele_Laino
    • one year ago
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    the subsequent drawing represents the situation of your exercise: |dw:1432832271358:dw|

  22. Michele_Laino
    • one year ago
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    this time: \[t = \sqrt {\frac{{2 \times 45}}{{9.81}}} \] is the time needed to Cleo and Clare to reach the base of their hotel. During that time they have to travel, horizonatally by a distance L=15 meters, in order to reach the swimming pool, and the time needed to travel that distance L, is: \[{t_1} = \frac{{15}}{v}\] where v is the horizonatl velocity of Cleo and Clear. Now we have to equate those two times, so we can write: \[\sqrt {\frac{{2 \times 45}}{{9.81}}} = \frac{{15}}{v}\] |dw:1432832762849:dw|

  23. Michele_Laino
    • one year ago
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    so, what is v?

  24. anonymous
    • one year ago
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    i dont understand the equation like how its written so i dont know

  25. Michele_Laino
    • one year ago
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    do you know the law of falling bodies?

  26. anonymous
    • one year ago
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    no

  27. anonymous
    • one year ago
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    would the answer be 3.03 seconds ?

  28. anonymous
    • one year ago
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    A bug and the windshield of a fast-moving car collide. Indicate whether each of the following statements is True or False (1 point each) • The forces of impact on the bug and on the car are the same magnitude. • The impulses on the bug and non the car are the same magnitude • The changes in speed of the bug and of the car are the same magnitude. • The changes in momentum of the bug and of the car are the same magnitude. • The accelerations of the bug and the car are the same magnitude.

  29. anonymous
    • one year ago
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    @Michele_Laino

  30. Michele_Laino
    • one year ago
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    please wait a moment

  31. anonymous
    • one year ago
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    okay :)

  32. Michele_Laino
    • one year ago
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    the answer to the first question is: \[v = \sqrt {\frac{{9.81}}{{2 \times 45}}} \times 15\]

  33. anonymous
    • one year ago
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    45=4.9t^2 9.184=t^2 t=3.03sec That is what i got for the first

  34. Michele_Laino
    • one year ago
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    yes I know, nevertheless, you have to compute the velocity v, since your problem asks for a velocity

  35. anonymous
    • one year ago
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    so it that right? and the 2nd one i got true, true, false, true, false?

  36. Michele_Laino
    • one year ago
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    we have: \[v = \sqrt {\frac{{9.81}}{{2 \times 45}}} \times 15 = 4.95\;m/\sec \]

  37. anonymous
    • one year ago
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    could you right the equation using the thing on here? thats the part that is getting me a little mixed up

  38. Michele_Laino
    • one year ago
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    I'm pondering....

  39. Michele_Laino
    • one year ago
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    the requested equation is: \[v = \sqrt {\frac{g}{{2 \times h}}} \times d = 4.95\;m/\sec \]

  40. Michele_Laino
    • one year ago
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    |dw:1432834613891:dw|

  41. anonymous
    • one year ago
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    ohhhhhh okay i think i get it, what about the second?

  42. Michele_Laino
    • one year ago
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    please, I'm pondering...

  43. Michele_Laino
    • one year ago
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    I think the first option, since we have to apply the third law of Newton, and the fourth option since for an isolate system the total momentum is conserved

  44. anonymous
    • one year ago
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    its asking which ones are true and which are false, what are you saying those are?

  45. Michele_Laino
    • one year ago
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    first option is true, second and third options are false, fourth option is true, fifth option is false

  46. anonymous
    • one year ago
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    Using 6400 km as the radius of Earth, calculate how high above Earth’s surface you would have to be in order to weigh 1/16th of your current weight. Show all work leading to your answer OR describe your solution using 3 -4 complete sentences.

  47. Michele_Laino
    • one year ago
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    here we have to apply the formula of Newton, about the interaction force between two object of masses M, and m respectively, which are separated by a distance d \[F = G\frac{{Mm}}{{{d^2}}}\]

  48. anonymous
    • one year ago
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    okay, so what goes where in the equation?

  49. Michele_Laino
    • one year ago
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    the weight W of an object whose mass is m, on the earth surface is given by the subsequent formula: \[\Large W = G\frac{{{M_T}m}}{{{R^2}}}\] where R is the earth radius

  50. Michele_Laino
    • one year ago
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    the weight of the same object, whose mass is m, at an altitude h with respect to the earth surface is given by the subsequent formula: \[\Large {W_1} = G\frac{{{M_T}m}}{{{{\left( {R + h} \right)}^2}}}\]

  51. Michele_Laino
    • one year ago
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    in both formulas M_T is the mass of the earth

  52. Michele_Laino
    • one year ago
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    now, we have to use this condition: \[\Large \frac{{{W_1}}}{W} = \frac{1}{{16}}\]

  53. Michele_Laino
    • one year ago
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    after a substitution, we get this condition: \[\Large {\left( {\frac{R}{{R + h}}} \right)^2} = \frac{1}{{16}}\] what is h?

  54. anonymous
    • one year ago
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    the height?

  55. Michele_Laino
    • one year ago
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    yes!

  56. Michele_Laino
    • one year ago
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    |dw:1432838068670:dw|

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