Recent content by Doubell

i wanted to know if it is right to say at state two at mechanical equilibrium the pressure P2 = Patm + (the Weight of the piston/area of the piston)?

Homework Statement A piston-cylinder device attached below is shown in Figure 6.A-31where Patm = 100 kPa Tamb = 150°C piston Ac = 0.01m2 mp = 100 kg water T1 = 350°C P1 = 400 kPa z1 = 0.5m Figure 6.A-31: Piston-cylinder device. The initial position of the piston is z1 = 0.5 m and the piston...

after re looking at the image in the problem statement i saw the indication ( air pressure , 'p'). so i guess instead of atmospheric pressure , the pressure P should be the variable in the equation. Pa - (900kg/m3*9.81m/s2*0.6m)- (P )+((0.6m+1.8m)*9.81m/2*900kg/m3)) = Pb . i would now have a...

Homework Statement An inverted U-tube monometer, as shown in Figure attached , has air at the top of the tube. If the pipes contain oil (s.g. = 0.9), h1= 0.6 m, h2= 1.8 m and h= 0.45 m, determine the difference in pressure between point B and point A. Homework EquationsThe Attempt at a Solution...

yes it is as u say but the problem still remains the same the∫e3x/x would still have to be determined

Homework Statement The question specifies the auxiliary equation given is (D^2 + D - 2) = (e^x)/(x) the method of variation of parameter must be used to find the particular solution to the right hand function. then finally the general soultion should be stated.Homework Equations variation of...

Homework Statement In 2-D space, the maximum number of unknown information is... Select one: a. 4 b. 2 c. 3 d. 1 e. 0 Homework Equations The Attempt at a Solution i think the answer is 1 comments would be appreciated

Homework Statement a man of mass 70kg rides a bicycle of mass 15kg at a steady speed of 4 ms-1 up a road which rises 1.0 m for every 20m of its length. what power is the cyclist developing if there is a constant resistance to motion of 20N Homework Equations (P= F*V) , ( p= W/t) The...

i apologize for the shouting.

thanks and i will remember no text speaking

ok for x= -1 ; (-1)^1/3 -4/(-1)^1/3 = 3 THIS IMPLIES -1 (-4)/-1 = -1 +4 = 3 AND FOR X = 64 ; (64^1/3) - 4/(64^1/3) = 3 WHICH IS 4 - 4/4 = 4-1 = 3 THE SOLUTIONS ARE TRUE FOR BOTH VALUES OF X. I WOULD LIKE TO SEE A DIFFERENT APPROACH THOUGH.

Homework Statement SOLVE FOR X IN THE EQUATION: X^(1/3) - 4(X^[-1/3]) = 3 Homework Equations The Attempt at a Solution I LET X^1/3 = Y THEN THE ORIGINAL EQUATION BECOMES Y - 4* 1/Y = 3 MULTIPLYING THE ENTIRE EQUATION BY Y RESULTS IN Y^2 - 4 = 3Y THEREFORE Y^2 -3Y - 4 = 0...

Writing my posts more clearly i noticed that my post are not as clear as u guys eg i write log2x when in ur posts its clear to understand any advice on how i can post my questions in a similar fashion as yours?

Homework Statement by substituting y = log2x solve for x in the following equation: √log2x = logs2√x Homework Equations logab=c then a^c = b The Attempt at a Solution if y = log2x then the equation becomes √y = log2 x^1/2 this implies √y = 1/2 log2x which simplifies to √y = 1/2 y...

ok so logarithms are not defined by negative values hence the root for the original equation would have to be the positive value of x which would have been {-2 +[28^1/2]}/2 is that the final solution.