the average heat transfer co-efficient over the entire length of the plate (ha) and the local heat transfer co-efficient (hl), in case of heat transfer over a flat plate in lminar zone is related as
ha = 2hl
heat lost by the vessel and water= [150 x 4.2 x (30 - 5) j + 100 x 0.4 x (30-5)] j = (150 x 4.2 x 25 + 100 x 0.4 x 25) j = (15,750 + 1000) j= 16,750 j heat gained by ice to change into water at 0 °c = m x specific latent heat of fusion of ice = {m x 336} heat gained by water to come to final temperature (5 °c) = m x specific heat capacity of water x final temperature = (m x 4.2 x 5) = 21m if there is no loss of heat to the surroundings, according to the principle of calorimetry: heat gained by ice to melt and come to final temperature = heat lost by vessel and water or 336m + 21m= 16,750 or 357m = 16,750 or m = 16750/357 = 46.92 Hi,


This defines the “local data area”. The local data area is an area of storage allocated to a job. You can see the contents of the *LDA by using DSPDTAARA *LDA, you can also change the *LDA using CHGDTAARA.


This used to be a popular way of passing data and parameters between programs on the System/36.


Regards,


Martin Gilbert.


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[The ILE RPG Reference also discusses the statement here](http://publib.boulder.ibm.com/infocenter/iseries/v5r4/topic/books_web/c0925086670.htm#HDRZZNAMVR).


Bruce Vining as per the question:
modal class = 35 -45
l = 35
h = 10
f1 = 23
f0 = 21
f2 = 14

= 

= 36.8
the above given data can be represented in the form of table as below:



= 35.37
the mode of the data shows that maximum number of patients in the age group of 36.8, whereas the average age of all the patients is 35.37. A food with higher specific heat most likely causes burns. solution: `ch\_(4)(g) + 2o\_(2)(g) to co\_(2)(g) + 2h\_(2)o(g) + "heat. (evolved)"` more like retain oxygen intake /efficiency. the formulas for converting between degree celsius and degree fahrenheit are: °f = (°c * 9/5) + 32 °c = (°f - 32) * 5/9 to find the temperature when both are equal, we use an old algebra trick and just set of = oc and solve one of the equations. °c = (°c * 9/5) + 32 °c - (°c * 9/5) = 32 - 4/5 * °c = 32 °c = - 32 * 5/4 °c = - 40 °f = (°f * 9/5) + 32 °f - (°f * 9/5) = 32 - 4/5 * °f = 32 °f = - 32 * 5/4 °f = - 40 so the temperature when both the celsius and fahrenheit scales are the same is - 40 degrees. given to prove 

then taking l.h.s.

we have 

                    





                                  

 

  

=  r.h.s.

hence proved. a) he helium (he) is an element by itself. he atom can only exhibit dispersion forces.b) hclh          1a                   1 × 1 e– = 1 e–cl         7a                   1 × 7 e– = 7 e–   total: 8 valence e–[readmore]cl is more electronegative than h, so hcl is a polar covalent compound. hcl exhibits dipole-dipole interaction.c) h2oh          1a                   1 × 2 e– = 2 e–o          6a                   1 × 6 e– = 6 e–   total: 8 valence e– since hydrogen is bonded directly to oxygen.h2o exhibits hydrogen bonding.d) cl2cl         7a                   1 × 7 e– = 14 valence e–composed of the same atom. there's no electronegativity difference.cl2 can only exhibit dispersion forces. generaloraverage rate is the change in the concentration of a compound over a period of time.the average rate the reactants and products in a reaction are equal when the coefficient for each is 1.the rate for the reactants are negative. this is because the reactants are being consumed in the reaction. the rate for the product c is positive because it is being produced in the reaction.for a reaction where the coefficients of the compounds are different, the rate is multiplied by the reciprocal of the coefficient.example: 2a + b → 3cthe given reaction is:3o2(g) → 2o3(g)­given the rate of o2 consumption is -2.17x10-5 mol/l∙snote that value is negative to indicate that it is decreasing as the reaction proceeds.rearranging the rate equation to give the reaction rate is terms of [o3]: