Recent content by Nyxious

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    Energy carried by electromagnetic waves question

    The cross sections area being the same as the area of the tilted surfaces width face. Was that the question you were asking? I may have read your question wrong.
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    Energy carried by electromagnetic waves question

    Cos (theta) = adj (Height of cross section) / hyp (length of surface), correct?
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    Energy carried by electromagnetic waves question

    Would the cross sections area be smaller than the area of the tilted surface?
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    Energy carried by electromagnetic waves question

    It's an equation given by my textbook. Malus's Law I believe? Any suggestions on other equations I could use?
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    Energy carried by electromagnetic waves question

    Hmm, both of you have stated a possible problem about the equation S2=S1 x cos^2 (theta) but I can't seem to find the problem. Doesn't this equation show how the intensity is decreased due to the angle of the axial tilt? I believe I'm missing something very obvious haha
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    Energy carried by electromagnetic waves question

    1. The power radiated by the sun is 3.9 x 10^26 W. The Earth orbits the sun in a nearly circular orbit of radius 1.5 x 10^11 m. The Earth's axis of rotation is tilted by 27 degrees relative to the plane of orbit, so sunlight does not strike the equator perpendicularly. What power strikes a...
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    What is the displacement vector needed to reach a research station on a safari?

    Okay, I went back and did everything again and here is what I got: C (or the distance remaining to get to the research lab) = 4.8120km The angle = 47.9° Can someone check to see if they got the same or if I'm still missing something?EDIT: Got it correct. Thanks for the help guys!
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    What is the displacement vector needed to reach a research station on a safari?

    I know, i said the way I did it did not work with getting the magnitude. I went back and found a magnitude of 8.269km which was incorrect, and then i tried subtracting the components and got 4.8120km, which is close to what i got when I just subtracted the final destination (6.49km) and the...
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    What is the displacement vector needed to reach a research station on a safari?

    Thanks for the help! I seem to be having another problem arise with my original magnitude I found. I see why 4.58km wouldn't work in this situation, but the program we use for homework is counting this correct. Any ideas why? I feel like my magnitude should be larger. Also, the angle to get...
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    What is the displacement vector needed to reach a research station on a safari?

    That was actually the first thing I did. That's actually how I kinda assumed the magnitude to be 4.58km. The angle however, is a mystery to me.
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    What is the displacement vector needed to reach a research station on a safari?

    Hello all! I've been struggling with this problem for a couple of hours and I just can't seem to wrap my head around on how to do it. Here it is: 1. On a safari, a team of naturalists sets out toward a research station located 6.49 km away in a direction 38.5 ° north of east. After traveling in...
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