Recent content by Jgoshorn1

So do you think it was just a typo that he wrote 2/x2+4 instead of 1/x2+4 because that 2 really confused me

How/where are you getting x/x2+4 and 2/x2+4 from?

Homework Statement ∫(x3+4)/(x2+4)dx Homework Equations n/a The Attempt at a Solution I know I have to do long division before I can break this one up into partial fractions. So I x3+4 by x2+4 and got x with a remainder of -4x+4 to be written as x+(4-4x/x2+4). Then I rewrote...

Homework Statement ∫e2xarctan(ex)dx Homework Equations From the table of integrals: #92 ∫utan-1udu = (u2+1)/2)tan-1-u/2 + c or #95 ∫untan-1udu = 1/(n+1)[un+1tan-1-∫ (un+1du)/(1+u2) , n≠-1 The Attempt at a Solution The answer is 1/2(e2x+1)arctan(ex) - (1/2)ex + C I don't...

So, what I've done is set q(in) = q(out) => -K1(dh/dl)=K2(dh/dl) => 6((h-1)/5) = 3((13-h)/5) => h(@x=5) = 5m Then I set the used set of boundary Conditions (BC) to interpret the solution to LaPlace in 1D (h=cx+D): For BC @ x=0 and h=1 =>1=c(0)+D => 1=D For BC @ x=5 and h=5 => 5=c(5)+1 => c=4/5...

Homework Statement Solve the Laplace equation in one dimension (x, i.e. (∂^2h)/(∂x^2)= 0) Boundary conditions are as follows: h= 1m @ x=0m h= 13m @ x=10m For 0≤x≤5 K1= 6ms^-1 For 5≤x≤10 K2 = 3ms^-1 What is the head at x = 3, x = 5, and x = 8? What is the Darcy velocity...

I suppose I can assume it is an ideal gas because all of the previous problems we have done have been dealing with ideal gases only. But even if it were implied, how would I find the number of moles of gas or the degree of freedom??

Homework Statement A heat engine does work by using a gas at an initial pressure of 1000 Pa and volume .1m 3. Step-by-step, it then increases the pressure to 10,000 Pa (at constant volume), increases the volume to .15m3 (at constant pressure), decreases the pressure back to 1,000 Pa (at...

Homework Statement An ideal gas in a cylinder occupies a volume of 0.065 m3 at room temperature (T = 293 K). The gas is confined by a piston with a weight of 100 N and an area of 0.65 m2. The pressure above the piston is equal to one atmosphere (atm = 1.013x10^5 Pa). The piston is free to move...

Thanks! I went back and re-did the volume and density of the block using 1.1kg as the weight of the water displaced and I got exact answers. I appreciate your encouragement!

My work looks correct and I don't know any other way to do it but I'm not getting it to match (or even come really close to) any of the other answers. Mass of block = 1.94kg = 1940g For density of block: d=m/v = 1940g/1020cm^3 = 1.90g/cm^3 Answers to chose from: [1.2 gm/cm^3, 1.8 gm/cm^3, 2.4...

Mass of water = 1.02kg = 1020 g Density of water = 1gm/cm^3 v(water)=m/d = 1020g/(1g/cm^3) = 1020 cm^3 = volume of water = volume of block? The options to chose from are [750 cm^3, 850 cm^3, 1100 cm^3, 1425 cm^3,2400 cm^3] The volume of the displaced water should = the volume of the block...

It's okay, I wasn't offended. I emailed my prof to figure out what's going on with this problem.. I don't care if I am getting it wrong or miss the credit for the problem.. I just want to know how to do it correctly! Until then I'm struggling with the second/third question: What is the...

I did that: Magnitude of buoyant force = weight of displace fluid Fb = m(water)g 10.012 = m(water)(9.81) 10.012/9.81 = m(water) = 1.02kg The problem is that 1.02kg wasn't a given option and I wanted to know if the way I was doing it was wrong or not.. Or if there is any other way of doing it.

Hi! So I worked on this problem for a very long time last night and I still wasn't getting an answer provided we can chose from [1.1 kg, 1.6 kg, 2.4 kg, 3.5 kg, 4.8 kg] The way I tried it was: Fnet=0 because it's in equilibrium I made a FBD with Force(gravity) pointing down, Force(buoyancy)...