Recent content by KAC

If I convert L to m3, it corresponds to Q, the flow rate of the fluid. So in this case, I don't actually need to solve for Q because this is already given? After conversions Q = 2.5*10-4; if I plug this into my equation: p= [(8nL)*Q]/pi*R^4 p=[(8*(1.0*10-3 Pa*s)*9m)*2.5*10-4 m3/s]/pi*0.01m4 p=...

The spring constant, k, has units of N/m; mg = F which is N (N = kg*m/s2), and U = N/m * m = N as well since it is the specific force in this case. So setting F = U in this case should be acceptable since the units are the same and U is the specific force that is working on the object, yes?

Two other formulas I may be able to use are Force of the spring = -kx (where x is displacement) and L(change in length) = mg/k. But both of these equations have two variables I do not know in them. When you say that spring energy varies as the square of the displacement, this leads me to believe...

Chet, I think that the units of velocity should be m/s. Since it is a fluid and not an object in this case, I am assuming that m^3/s is suitable to use?

Chet, I used U = F as the force; the equation for U that I used is the potential energy of the spring. Since it is being pulled down and there is no normal force in this case, only the weight force is acting on the object which is = mg. (mass*gravity).

Homework Statement A spring scale hung from the ceiling stretches by 6.3 cm when a 1.2 kg mass is hung from it. The 1.2 kg mass is removed and replaced with a 1.4 kg mass. What is the stretch of the spring? F=mg U (spring force) = 1/2kx^2 Homework Equations The units of cm need to be...

Homework Statement Water flows at 0.25 L/s through a 9.0-m-long garden hose 2.0 cm in diameter that is lying flat on the ground. The temperature of the water is 20 ∘C. What is the gauge pressure of the water where it enters the hose? Side question: does the velocity of the water flow need to...

So then I should set it up with the same equation Q = Q(specific heat) + Q(phase change)?: Q = McT + MLf Q = M(cT + Lf) M = Q/(cT + Lf) M = 2000J/(128J/kg*K * 303K) + 2.5*10^4 J/kg And with this work I did get the correct answer, M = 0.0314 kg; M = 31.4 g. Thanks again, this will help on my...

THANK YOU for pointing that out. I think I'm crunching numbers so much I really just am throwing them together.. but yes, I should have plugged 128 for c; I mistakenly multiplied the change in temperature by the heat of fusion and added that to the specific heat when I should have done it the...

The way that I computed it, is that since both equations had M in the formula, I would then have 2M after adding the formulas together, and after completing the work I'd need to divide my answer by two to account for this.. But now I see where that doesn't make sense! Since I'm taking out a...

Okay, so what I am thinking now, since 2000J is only given to us once, is that Q = McT + MLf. So, Q = McT + MLf; Q = 2M(cT + Lf); 2M = Q/(cT + Lf). When calculating this out I get: 2M = 2000J/[(128J/kg*K * (328-25)K) + 2.5 * 10^4 J/kg]; 2M = 0.03136kg; 0.03136/2 = 0.0157 kg or 15.6g would be...

I am trying to solve the problem.. But I feel like I'm going in the completely wrong direction with the way I've set up my problem solving, so I would appreciate the assistance of pointing out one part of my work that I might have calculated wrong or an equation that I shouldn't be using at all...

So for this, how could I set it up? Should I add the two products together or set the two equations equal to each other such as cT/Q = Lf/Q?

Homework Statement What is the maximum mass of lead you could melt with 2000 J of heat, starting from 25 ∘C ? Lead melts at 328∘C , its specific heat is 128 J/(kg⋅K) , and its heat of fusion is 2.5×10^4 J/kg . Homework Equations Need to find both the mass in the heat of transformation (Q =...

TSny, Thank you for your welcome! And thank you for pointing out my white mistake. LOL. So let me try this out again: v = 5km/h*1000m/1km*1h/3600s= 1.39 m/s. e = output/input; efficiency*Ecarb = KE; .25*368720J; KE = 92180J. P = K/time interval; t = K/P; t= 92180J/380W; t = 242.58 s...