Recent content by AbbeyC172

It is the latter, (V2^2 - V1^2)/2

Homework Statement I am trying to figure out how to get the V2^2 alone and I feel that this is very basic rearranging of equations but I have been stuck on it forever now. Thank you so much in advance. -75 kj/s= 4.09 kg/s (-190.3 +...

Ah I get it now! So could I just put in 0.0625 in instead of the 2lbm?

Thank you for reply! Lbs of what? All I found was a lb mole is 12 lbs of 12C. I’m not going to lie. Moles confuse the heck out of me.

Homework Statement A 3-ft^3 container is filled with 2-lbm of oxygen at a pressure of 80 psia. What is the temperature of the oxygen?Homework Equations pV= nRT T= PV/nR R= 10.7316 psia x ft^3/ lbmol x R The Attempt at a Solution Hi everyone! So I understand how to use the Ideal Gas Law but my...

Thank you for all the help! I will let you know what my professor says when he grades the paper.

So I worked with a tutor and we figured out that Fn= mg =100 lbs F= Ffriction = 20 lbs Force of friction = 20 Force of the man on the box: 20 lbs Work on box by the man: 20 (100) (1) = 2000 ft x lbs Total force = 0 Work being done on the system is zero because W = 0 (100) (1) = 0 So work done...

Ahh, the cosine for 0 degrees which is what we have on a horizontal plane is 1. The cosine would be zero at 90 degrees. The mass of the man: after converting.. m = f/a = 800.68/9.8= 81.7 I'm sorry, I think I'm getting more confused. I'm really lost as to what I'm even trying to figure out...

Cosine should be zero since we have no angle. Fman on block, x: 180 lbf Fblock on man, x: 180 lbf Ffloor on block, x: mg = 100 lbf(9.8) = 980 N Ffloor on man, x: = mg = 180 lbf(9.8) = 1764 N In the vertical direction FEarth on man, y: 180 lbf FEarth on block, y: 100 lbf Ffloor on block, y: 100...

This is tremendously helpful! Thank you. So since the velocity is constant, then the acceleration must be zero. But in regards to what the problem asks: am I on the right track to getting the force needed for the calculations? F(n)= m x a = F(n) = (81.65 kg) x 9.8 m/s^2 = 800.17 N Then...

It would be zero? Now I am so confused.

Normal force? 100 lbf because that is the weight of the crate and the floor exerts the same back on it? Wouldn't those forces cancel out? If using Fnet=ΣFapplied=ma, then could I convert the 180 lb man into 81.65 kg and multiply that by 9.8m/s^2?

1. The first F would be the total force, so F total = μ (F, force of box?) (9.8 m/s^2) so Ft = 0.2 (100) (9.8m/s^2) ? 2. Would the net force be equal to the force of the man minus the coefficient of friction? Thank you for the tip about the symbols. If you can't already tell, haha I'm quite new...

I would think the friction would have to be smaller than the net force in order to move the box?

The force of the friction is pushing back on the force from the man. The total force has to always be bigger than the force of friction. To calculate the force on the block, could I use: F= (F)g ? Because once I figure out the force on the block and the force on the man, couldn't I add them...