the value of head loss due to friction using chezy formula with constant c= 60 for a fluid, flowing &#8230;.
the value of head loss due to friction using chezy formula with constant c= 60 for a fluid, flowing through a 30 cm diameter and 70 m long pipe. velocity of fluid inside the pipe is 5 m/s.
the value of head loss due to friction using chezy formula with constant c= 60 for a fluid, flowing through a 30 cm diameter and 70 m long pipe. velocity of fluid inside the pipe is 5 m/s.
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 class intervals
frequency
(fi)
xi
di = xi - a
fi di


0 - 20
17
10
-40
-680


20 - 40
28
30
-20
-560


40 - 60
32
50 = a
0
0


60 - 80
p
70
20
20p


80 - 100
19
90
40
760


total
Î£
fi = 96 + p


Î£ fi di = 20p -
480


![](https://s3mn%20mnimgs%20com/img/shared/discuss_editlive/5130830/2013_07_08_13_01_36/mathmlequation8813644107607516415%20png) v = (4/3)πr³ 

dv/dt = 4πr².dr/dt

so, when the diameter is 20 cm we have:

-4π = 4πr².dr/dt

hence, dr/dt = -1/r²

i.e. dr/dt = -1/10² => -1/100

now, d = 2r

so, d(d)/dt = 2.dr/dt

=> d(d)/dt = -1/50 cm/hour

or, -0.2 mm/hour

:)> coefficient of friction is defined as the ratio of the force of friction acting between two objects and the force which is pushing them together. the value of the coefficient of friction varies from zero to one. there are two types of the coefficient of friction: static coefficient of friction kinetic (also known as sliding) coefficient of friction. therefore, the coefficient of friction depends on the nature of the surface. u (initial velocity) = 18 km/hr 
t ( time )= 2.5 sec
v( final velocity) = 0 m/s
from, the first equation of motionv = u +  a×t0 = 5 +  a×2.5a = - 2 m/s2 d = 21 m
r =![](https://img-nm%20mnimgs%20com/img/study_content/curr/1/7/3/41/1541/chapter%2011_html_96a2a55%20gif)
circumference = 2πr =![](https://img-nm%20mnimgs%20com/img/study_content/curr/1/7/3/41/1541/chapter%2011_html_521eed1b%20gif)= 66 m
length of rope required for fencing = 2 × 66 m = 132 m
cost of 1 m rope = rs 4
cost of 132 m rope = 4 × 132 = rs 528 let the speed of train = x km /hlet the speed of car = y km/hcase 1 : 120 km by train and 480 by cartime taken to travel 120 km by train = 120x hrstime taken to travel 120 km by car = 480x hrsaccording to first condition :120x + 480y = 8⇒1x + 4y = 115     
1case 2 : 200 km by train and 400 by cartime taken to travel 200 km by train = 200x hrstime taken to travel 400 km by car = 400x hrsaccording to second condition :    200x + 400y = 82060 = 813 = 253         as, 20 min = 2060 hr = 13 hr⇒1x + 2y = 124     
2subtracting 2 from 1, we get      2y = 115 - 124⇒2y = 8-5120⇒y = 80now, from 1, we get    1x + 480 = 115⇒1x = 11 - 120 = 160⇒x = 60so, speed of train = 60 km/hspeed of car = 80 km/h miles/hour is just that.
 
(3/4)/(1/12) = (3/4)(12) = 9 mph sylvia frank chlh 274 homework problems chapter 3 1. a. crude death rate: total populace death in a specific span-time at the mean year. b. age-specific rate: total incidence occurred in a specific-age group of the populace in a specific span time. c. cause-specific rate: total death of a populace from a specific-cause in a specific span-time at the mean year. d. proportional mortality ratio: total death within a populace due to a specific-cause divided by the total death in the populace e. maternal mortality rate: incidence of death because of birth of a child or due to pregnancy associated complications. f. infant mortality rate: rate of event of newborn death over a year by birth of newborn over a same interval. g. neonatal mortality Max 

Hello Max,


There is a little typo on the solution. In a warmer winter you will have gas consumption reduced by 10%, that implies a reduction of 10% of sales and variable costs (the hypothesis is to consider them linearly dependant).


Talink about the Infrastructure costs, they are composed by fixed (60%) and variable(40%) costs. Applying the saving on the variable part you have:


240*40%*10%=9.6  

  

So that the new Infrastructure costs will be 240-9.6=230.4  

  

Does it make sense?  

Feel free to contact me if you have any doubt.  

  

Luca