jasoncurious

Thanks. So, the "head" concept is something like a "conservation of height" thing?

Hi guys, I am currently studying fluid mechanics. This is my first time doing the subject. I've been reading some books. Then I realised something: 1. Do we always have to use an annulus as a fluid element? What's so special about the ring? Was it because it's easier to analyse? 2. Regarding "head" (dimension L), is it easier to understand by expressing it in L? Thank you.

Hi there! As long as I can remember, an addition reaction consists of both an electrophile and nucleophile. Everytime, the nucleophile is "the one attacking" and the electrophile is "the one attacked". If nucleophilic addition reaction is named in such a way that the reaction is initiated by a nucleophile (it attacks), then electrophilic addition reaction must be occurring in such a way that the electrophile is the one that initiates the reaction. My question is, how does an electrophile initiate an addition reaction when it's always "attacked"? Thank you.

I saw my friend solving it using Bernoulli equation, guess everything is possible in mathematics

I thought the d.e has to be in the form of dy/dx+P(x)y=Q(x) for integrating factor to be implemented?

y=vx dy/dx=v+x dv/dx vx(v+x dv/dx)=x^3+(v^2 x^2)/x v^2*x+vx^2 dv/dx=x^3+v^2*x Eliminate the v^2*x: vx^2 dv/dx=x^3 Divide both sides with x^2: v dv/dx=x vdv=xdx Continue the integration: y^2=x^2(x^2+c), where c is a constant

So, even if the differential equation is not homogeneous, we can use y=u*v as well? This is howthe solution starts: yy'=x^3+(y^2)/x Let y=vx dy/dx=v+x dv/dx Then the rest is substitution.

Thank you, I think I was caught in the thought that "only homogeneous differential equations can be solved".

Convert the following differential equation into separable form by using suitable substitution: yy’=x3+(y2/x) When I try to prove it’s being homogeneous or not, I found that yy’=x3+ y’= (lx)4+(ly)2/ (lx)(ly) Factorise it: (l^2x2)+(y^2) / xy Is this considered homogeneous? the l stands for lambda

Sorry, but I can't help but wonder, isn't that every thermodynamic situation involves entropy?

Hi guys, happy holidays. I need some help in choosing the usage of H (enthalphy) and U (internal energy) when dealing with thermodynamics. Any guidelines?

So, the black one is the piston. Since it is weighted, the pressure produced is lost in pushing the piston upwards?

100g of CO are contained in a weighted piston-cylinder device. Initially, the CO is at 1000kPa and 200oC. It is then heated until 500oC. Determine the final volume of the CO treating it as an ideal gas. This is what my lecturer gave in his solution: V2=(mRT2)/P=(0.1kg*0.2968kPa.m3/kg.K*(500+273)K)/1000kPA My question is: why is the original pressure (1000kPa) been used instead of the final pressure? Thanks for your help.

Thanks, so I assume this type of to be contextual?

Hi all, there's this function: y=(x^2-6*x+5)/(x-1) which can be reduced to y=(x-5)(x-1)/(x-1) and then to y=x-5 The question asked me to determine the domain and range of this function. Here's the problem, when the function is y=(x^2-6*x+5)/(x-1), the domain is every real number except 1. But when it's been reduced to y=x-5, the domain should be every real number. However, the answer stated that the domain is still every real number except 1. How should I solve this problem?