Recent content by ShineyPenny

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    Can Internet Security Ever Match the Safety of Physical Systems?

    From my Naval Systems Security course: "The best computer protection is only as good as its user."
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    Frequency, and finding wave speed. Professor mistake?

    For problem one, I notified the professor and he said the guide is incorrect...the answer is 34. Now it all makes sense for that one! For problem two, whoops >.< I take it I only use V=Lambda*Period and f=1/Period if I am only given the period and I'm trying to discover Velocity or...
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    Frequency, and finding wave speed. Professor mistake?

    Sure thing 1. Sorry...slight mix up in answers, the sheet says .029 but I am getting .0029. 1 / 340 = .0029. Am I even doing the right formula for this? 2. OK, so it is 16. Why am I using V=Lambda x Period and not V=Lambda / Period?
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    Frequency, and finding wave speed. Professor mistake?

    I am studying for an upcoming test, and it is known that there are hiccups in our study guide. I am stuck on two problems. Homework Statement 1. What is the frequency, in Hz, of a sound wave (v=340 m/s) with a wavelength of 10 m? 2. A series of ocean waves, each 8 m from crest to...
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    Estimating image distance in lens and mirror

    OK I have figured it out, the book is wrong and it was verified by the professor :) I inputted: (1/16f) + (1/8o) = .1875, then (1/.1875)= 5.33.They did not use a negative sign.Thank you! =)
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    Estimating image distance in lens and mirror

    I have no idea where 45 and 16 came from, sorry it was a long night :blushing: I input (1/8o) + (1/-16i) = .0625f, then go (1/.0625f) and come out with 16. Either I'm missing something big, or I haven't drank enough coffee this morning. I even plugged the numbers into this website...
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    Estimating image distance in lens and mirror

    OK, I believe I have it after running a search for it...it's actually not in the book :(Is the equation: (1/Object Distance)+(1/Focal Length) = (1/image location) (1/45) + (1/36) = .05; then (1/.05) = 5.999 This is it? I'm just making sure :)
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    Estimating image distance in lens and mirror

    I'm having difficulty understanding how to estimate the location of an image in a lens/mirror. I tried understanding it by going off the books question/answer section: "A spherical concave lens has a focal length of 16 cm, and an object is placed 8 cm from the lens." It says the answer is...
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