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I typed the problem word for word, I'll include the multiple choice answers in this post: a) 3 b) 15/4 c)4 d)9 e)12

Homework Statement If theta increases at a constant rate of 3 radians per minute, at what rate is x increasing in units per measure at the instant when x equals 3 units? The Attempt at a Solution I drew the triangle and I came to the conclusion that I needed to use 5sin(theta)=x...

Homework Statement Need to find the derivative: y=logx/2-logx=logx*(2-log(x))^-1 The Attempt at a Solution The derivative of log(x) is 1/xln(10) and the derivative of (2-log(x))^-1 is -(2-log(x))^-2*1/xln(10)? This ended as 1/xln(10)(2-logx) - logx/xln(10)(2-log(x))^2...

So how did the numerator become: (2-cos(3x)-cos(4x)) = (1-cos(3x)) + (1-cos(4x))?

lim as x-> 0 (2-cos3x-cos4x)/(x). I'm not sure how the numerator became (1-cos3x)(1-cos4x)/(x) What am I missing? Could someone please point it out? Is it a trignometric factoring formula I'm not thinking of?

Homework Statement limi as x-> positive infinity x+r((x^2)+2x) The Attempt at a Solution multiply by conjugate x-r((x^2)+2x) I get (x^2)-(x^2)+2x/x-r((x^2)+2x) Which becomes 2x/x-r((x^2)+2x) Which I end up with 2/1-r(1-(2/x)) But I go wrong somewhere because I end up...

http://math.colorado.edu/~jkeller/math1300/lectures/L8limits3.pdf [Broken] Example 8, the 2nd to 3rd step... where did the negative go? It's (x^2 - 3x) - x^2 then it became 3x in the numerator. Where'd the negative go? I don't see any other sign changes so could someone please tell me?

Ah I understand it now lol

For the first integral no I did not.

I'm confused as to when to change the limits on a definite integral. Ex. Integral with the limits a=1, b=5, 3/(x+1)dx I set u = x+1 and du = dx I used u-substitution and everything worked out fine. However for this one... Ex. Integral with the limits a = 0, b = 2...

There are only 2 graphs: f'(x) and g'(x). I understand the one for g(x) now however... For the f'(x) graph, f'(x) is always above the x-axis. That means the slope is always positive?

Homework Statement When given the graph of f '(x) the graph sort of oscillates above the x-axis and the graph of g '(x) which starts in Quadrant III sort of curves up, goes through point (0,0) keeps going up then begins to even out. Homework Equations How can you tell how many...

Thanks again!

First we fired at 0 degrees, which we then used to find Vo Vix and Viy were found after he gave us the angle of degree to fire it at. For Delta x we had a Delta x for when it was 0 degrees. We then were given an angle to fire at. We have to find the Delta x for when it will be fired at 12...

Fired at 0 degrees: Delta x: For 1 click: 1.28 meters For 2 clicks: 1.68 meters For 3 clicks: 2.28 meters Note* clicks means how much we compressed the spring within the cannon Delta y = 0.87 meters Each of the three rings are set at equal intervals of 1/4 of delta x.

Homework Statement Part 1 We fired a cannon 0 degrees to the horizontal line and had to calculate Vo. We then measured Delta x and Delta Y. Part 2 We were then given a degree in which to fire the cannon at, which was 12 degrees. We then had to calculate the projectile's path through...

I tried a different way by doing: Delta y = Viyt + (1/2)(g)(t^2) -56 = -4.9t2 t= 3.3 Then: Delta x = Vot + (1/2)axt2 45 = Vix(3.3) Vix = 13.63 Is this correct now? I think it is. ----------------------------------- Also would you mind if I asked you about another problem...

Is it plausible to first find time then use the equation: Delta x = Vixt + (1/2)at2 Since ax = 0, the second half of the equation would be gone then right? So I could find Vix = Delta x/t And I think Vix = Vo, that would be my answer?

Since the ball is thrown horizontally that means Vo = Vix right? Since the Viy = 0 because there is no vertical velocity. This also means that (theta) = 0 degrees right? But when I try to find Viy I get: Viy = Vosin(theta) Vo = 0/sin(0) But that's undefined since sin(0) is 0 and you...

Oh right how did I miss that... So since it's thrown horizontally, theta = 0 right? So then Viy = 0 Then use the equation: Delta y = Viyt + (1/2)gt2 Delta y = (0)(1.6) + (1/2)(-9.8)(1.6)2 Delta y = -12.544 However since it's a building, the answer would be +12.544 right...

Vo = 22 m/s as well? I would have to calculate theta too wouldn't I? I tried an equation: Vfy = Viy+ayt 0 = Viy + (-9.8)(1.6) Viy = 15.68 m/s Delta y = Viyt + (1/2)gt2 Delta y = (15.68)(1.6) + (1/2)(-9.8)(1.6)2 Delta y = 25.088 - 12.544 Delta y = 12.544 meters (The answer some...

Problem #2 Is the time I calculated wrong too? Or is it just the second part of my calculations that is wrong? So to find Viy I should do: Vix = Vo cos(theta)?

I don't know if my calculation for time is right though.

Homework Statement Problem #1 A ball is thrown horizontally from the roof of a building 56m tall and lands 45m from the base. What was the ball's initial speed. Problem #2 A ball thrown horizontally at 22.2m/s from the roof of a building lands 36.0 meters from the base of the building. How...

Oh I understand now. However there is a slight problem, we are not told the radius or diameter of the ring on the ring stand. The only measurements we have is the Delta Y, and degree of cannon fire.

I have a graph of f and a graph of f'. That's what I meant by graphs. Sorry if it wasn't clear. Thanks though!

Homework Statement What is the relationship with a function's rising, falling, high point or low point to it's derivative? The Attempt at a Solution I have plotted my graphs, I can see that they intersect at the high and low points. But what is the relationship Also on another...

radius of...? what? When I use the equation Y = (1/2)ayt2+Viyt + Delta Y. I calculated that the first ring stand is 0.5075 meters away from the counter we launch the cannon off of, and that it takes0.128825 seconds for the projectile to reach it. So what I did was: y =...

Homework Statement For my lab, I need to calculate the height of the ring stands so the projectile can pass through them; as well as the angle the projectile will come at them. I have calculated/measured the following things: Muzzle Velocity: 4.11m/s Vix= 4.02m/s Viy= 0.85m/s Delta Y: 0.87...

Oh ok thanks all.

v initial y...? what?

Homework Statement My group and I had a lab, and we need to calculate time in order to proceed with the lab. We need to calculate how far and how long the projectile will travel using the degree we are given to hit the target that will be placed X meters using our calculations. We have a...

Almost 7 years later and we just did this problem in class! Still useful.

One question though: Are these ideas capable of a High School Honors Physics student?

I'm just getting really frustrated because I keep confusing myself for some reason. Thanks for the help though, much appreciated.

Since the SI for time is seconds then, if it's 600 ps (pectoseconds right?) what else is there to convert to?

I said it was wrong initially because the way I derived it was wrong.

Ok I got it. My trouble is that I'm not sure what to convert it into. Like if it's mg I have to convert it go grams right? Or if it's pectoseconds I convert it to a minute? If fentometers then just to meters?

it's a sheet of paper. I got this one now though. 86.6/1000 = 0.0866 which is 8.66E-2. Except now I'm getting stuck on 45 microVolts. I don't know how to convert this one. What do you convert microVolts into? Volts?

Sorry for the typo. I meant 86.6mm.

Homework Statement Convert the following as full (decimal) numbers with standard (base) SI units: 86.6 m The Attempt at a Solution I thought I could just change it to 8.66x10^1 then just add the 10^-3 because of the milli to make it 8.66x10^-2. But I'm wrong so could someone help me?

lots of ideas, thanks everyone! Does anyone else have any others?

Hi, I am having a hard time figuring out an idea I can use as my senior project in High School. I am currently in an Honors level physics class and the school year has just begun so we haven't learned much yet but. I am very interested in physics. So if anyone could help me with some ideas it...