Scott Hartnell's Mane Posted February 21, 2013 Share Posted February 21, 2013 in my school days we didn't have CDC to help us with homework and get wrong answers, we did our homework drunk or high to get wrong answers instead. I did my homework in high school stoned off my ass and graduated with a 3.9. Drunk may be apropos in this statement, but pot causes minimal effect on the brain. People have been believing myths demonizing it for years. THC is actually shown to improve concentration and as it is a neuroprotective...this means it actually protects against brain damage, especially against alcohol induced brain damage. Link to comment Share on other sites More sharing options...
D.Doughty Posted February 22, 2013 Share Posted February 22, 2013 electrons are accelerated from rest through a potential of 500 v in an electron gun. Assuming that all of the work done on the electrons produce kinetic energy, what is the speed of the electrons leaving the gun Help please! mass of electron=9.1x10^-31 electron charge= 1.6x 10^-19 Link to comment Share on other sites More sharing options...
Kass9 Posted March 4, 2013 Share Posted March 4, 2013 Wow, is that for POLI 100? I guess Erickson uses the same questions every year As for the question itself, you could look into Switzerland and their independence from the EU and argue that sovereignty is still significant. Canada's protests against NAFTA? You could argue that sovereignty is necessary because states have many different interests. Take a look at Samuel Huntington's article "Clash of Civilizations." I did a different question last year so yeah...don't know if that even helps. Hmm, I see. I am actually arguing for globalization. Thanks for the insight because now I can structure something to counter that. Link to comment Share on other sites More sharing options...
D.Doughty Posted September 7, 2013 Share Posted September 7, 2013 On a straight road, a car speeds up at a constant rate from rest to 20 m/s over a 5 second interval and a truck slows at a constant rate from 20 m/s to a complete stop over a 10 second interval. How does the distance traveled by the truck compare to that of the car? A) There is not enough information to answer the question. B)The truck travels half as far as the car. C) The truck travels twice as far as the car. D)The truck travels the same distance as the car. Car I know vi=0 vf=20 t=5 vf=vi+at a=4 vf^2=vi^2+2ad d=50 truck vf=20 vi=10 t=10 d=? vf=vi+at a=-2 vf^2=vi^2+2ad d=100 anyone could tell me if i did it right? d=? Link to comment Share on other sites More sharing options...
n00bxQb Posted September 7, 2013 Share Posted September 7, 2013 On a straight road, a car speeds up at a constant rate from rest to 20 m/s over a 5 second interval and a truck slows at a constant rate from 20 m/s to a complete stop over a 10 second interval. How does the distance traveled by the truck compare to that of the car? A) There is not enough information to answer the question. B)The truck travels half as far as the car. C) The truck travels twice as far as the car. D)The truck travels the same distance as the car. Constant rate of acceleration means that you can just take an average speed. (20+0)/2 = 10 m/s over 5 seconds = 50 metres (0+20)/2 = 10 m/s over 10 seconds = 100 metres C) Truck travels twice as far as the car Link to comment Share on other sites More sharing options...
D.Doughty Posted September 9, 2013 Share Posted September 9, 2013 A diver jumps off a 3.00 meter high diving board with an initial velocity of 1.75 m/s vertically upward. a) What is the diver’s velocity when she reaches the water? (Assume the surface of the water is 3.00 meters below the board.) What is the highest point the diver reaches above the water? is this projectile motions? could someone show me how to do it i tried dy=3 viy=1.75 vfy=0 a=9.8 t=? d=vit+1/2at^2 then using quadratics i got t=0.62s am i on the right track? Link to comment Share on other sites More sharing options...
kazin! Posted September 9, 2013 Share Posted September 9, 2013 A diver jumps off a 3.00 meter high diving board with an initial velocity of 1.75 m/s vertically upward. a) What is the diver’s velocity when she reaches the water? (Assume the surface of the water is 3.00 meters below the board.) What is the highest point the diver reaches above the water? is this projectile motions? could someone show me how to do it i tried dy=3 viy=1.75 vfy=0 a=9.8 t=? d=vit+1/2at^2 then using quadratics i got t=0.62s am i on the right track? Haven't done physics in a while but here's my crack at it... Part a) vo = 0 m/s a = 9.8 m/s^2 xo = 3 m x = 0 m v^2 = vo^2 + 2a(x-xo) v^2 = 2(9.8)(3-0) v = root(2*9.8*3) v = 7.67 m/s But....I am not sure because it doesn't give you how high the diver jumped. Link to comment Share on other sites More sharing options...
D.Doughty Posted September 9, 2013 Share Posted September 9, 2013 Haven't done physics in a while but here's my crack at it... Part a) vo = 0 m/s a = 9.8 m/s^2 xo = 3 m x = 0 m v^2 = vo^2 + 2a(x-xo) v^2 = 2(9.8)(3-0) v = root(2*9.8*3) v = 7.67 m/s But....I am not sure because it doesn't give you how high the diver jumped. That's part b, the highest part. I thought it was parabolic so we would separate into x and y dimensions no? Link to comment Share on other sites More sharing options...
LeanBeef Posted September 9, 2013 Share Posted September 9, 2013 need HELP quick. x-13/13x100=0.008 how do i find x? Link to comment Share on other sites More sharing options...
D.Doughty Posted September 9, 2013 Share Posted September 9, 2013 need HELP quick. x-13/13x100=0.008 how do i find x? Multiply 0.008 by 1300. Then isolate x by adding 13 Link to comment Share on other sites More sharing options...
LeanBeef Posted September 9, 2013 Share Posted September 9, 2013 Multiply 0.008 by 1300. Then isolate x by adding 13 thanks i thought about it way more than i should have Link to comment Share on other sites More sharing options...
LeanBeef Posted September 9, 2013 Share Posted September 9, 2013 Multiply 0.008 by 1300. Then isolate x by adding 13 wait how did you get 1300? Link to comment Share on other sites More sharing options...
kazin! Posted September 9, 2013 Share Posted September 9, 2013 That's part b, the highest part. I thought it was parabolic so we would separate into x and y dimensions no? Perhaps. Can you just ignore that added height/time from the jump then? I'll try part b tomorrow. Link to comment Share on other sites More sharing options...
D.Doughty Posted September 9, 2013 Share Posted September 9, 2013 wait how did you get 1300? 13 x 100 Link to comment Share on other sites More sharing options...
D.Doughty Posted September 15, 2013 Share Posted September 15, 2013 given that f (x)= square root [1+(lnx)^2] for x greater or equal to 1, find the inverse function how do you do this? x=root 1+ (lny)^2 x^2=1+(lny^2) x^2-1=lny^2 ln root (x^2-1)=y what am i doing wrong? Link to comment Share on other sites More sharing options...
D.Doughty Posted September 18, 2013 Share Posted September 18, 2013 Find the appropriate trigonometric function to describe the following rhythmic processes: Sleep-wake cycles with peak wakefulness (W = 1) at 8:00 am and 8:00pm and peak sleepiness (W = 0) at 2:00pm and 2:00 am. the answer is this: W(t) = 0.5 + 0.5 cos[π(t − 8)/6]. anyone who why you need to include the bolded part. i get every other part Link to comment Share on other sites More sharing options...
Cunuck Posted September 18, 2013 Share Posted September 18, 2013 Find the appropriate trigonometric function to describe the following rhythmic processes: Sleep-wake cycles with peak wakefulness (W = 1) at 8:00 am and 8:00pm and peak sleepiness (W = 0) at 2:00pm and 2:00 am. the answer is this: W(t) = 0.5 + 0.5 cos[π(t − 8)/6]. anyone who why you need to include the bolded part. i get every other part If you try plugging in t = 8 (8:00 AM), you notice that without the added 0.5, W(8) = 0.5. But in the question it states that at 8:00 AM you're at peak wakefullness (W = 1). That's just a way to check the requirement of the 0.5 part. I'm not the best explainer Link to comment Share on other sites More sharing options...
D.Doughty Posted September 18, 2013 Share Posted September 18, 2013 If you try plugging in t = 8 (8:00 AM), you notice that without the added 0.5, W(8) = 0.5. But in the question it states that at 8:00 AM you're at peak wakefullness (W = 1). That's just a way to check the requirement of the 0.5 part. I'm not the best explainer But how would you form that equation without knowing it? Link to comment Share on other sites More sharing options...
Cunuck Posted September 18, 2013 Share Posted September 18, 2013 Okay so you know the following values from the question: W(8) = 1 W(20) = 1 W(2) = 0 W(14)=0 So one thing you automatically know is the amplitude of the function ((max-min)/2) which is 0.5. Also, you know that W(t) never outputs a negative value since the minimum value is 0. Therefore we know that our function must be of the form: x + 0.5cos(blah) where x is some number. Let me know if I'm making any sense before I continue further Link to comment Share on other sites More sharing options...
D.Doughty Posted September 18, 2013 Share Posted September 18, 2013 Okay so you know the following values from the question: W(8) = 1 W(20) = 1 W(2) = 0 W(14)=0 So one thing you automatically know is the amplitude of the function ((max-min)/2) which is 0.5. Also, you know that W(t) never outputs a negative value since the minimum value is 0. Therefore we know that our function must be of the form: x + 0.5cos(blah) where x is some number. Let me know if I'm making any sense before I continue further yup you are i understand why its 0.5 cos[π(t − 8)/6] but i dont understand why its 0.5 cos[π(t − 8)/6]+0.5. Why vertical shift by .5? Link to comment Share on other sites More sharing options...
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