The water of a 14' times 48' metal frame pool can drain from the pool through an opening at the side of the pool. the opening is about h = 1.2 m below the water level. the capacity of the pool is v = 3710 gallons, the pool can be drained in t = 9.3 mins. p_0 is the pressure of the atmosphere, rho is the density of the water. (a) write the bernoulli's equation of the water at the top of the pool in terms of p_0, rho, h. assuming the opening is the origin 0% deduction per feedback. (b) write the bernoulli's equation of the water at the opening of the pool in terms of p_0, rho, h and v, where v is the speed at which the water leaves the opening. assuming the opening is the origin. (c) express v^2 in terms of g and h. (d) calculate the numerical value of v in meters per second. (e) express the flow rate of the water in terms of v and t. (f) express the cross-sectional area of the opening, a, in terms of v, v and t. (g) calculate the numerical value of a in cm^2

  1.the balanced   chemical equation is as below

cu +   2 ag(no)₃ → cu(no₃)₂   +2 ag

2.the   much silver   produced   is   50.9 g

step 1: find the   moles of cu

moles =   mass÷   molar mass

  from periodic table the molar mass of cu = 63.5 g/mol

moles is therefore = 15.00 g÷ 63.5 g/mol = 0.236   moles

step 2: use the mole ratio to determine the   moles of ag

from equation above   cu: ag   is 1: 2

therefore the moles of ag   =0.236   moles   x 2/1 =0.472   moles

step 3: find the mass   of ag

mass = moles x molar mass

from periodic table   the molar mass of ag =107.87 g/mol

mass   = 0.472   moles x 107.87 g/mol = 50.9 g

3.   the   much   silver   collected worth is   1.797 oz

  convert   50.9 g   to oz

that is   1 gram = 0.0353 oz

          50.9 grams =?   oz

by cross multiplication

={(50.9 grams x 0.0353 oz) / 1   gram}   =1.797 oz

grav forces affect all objects magnetic forces affect magnetic objects and electric currents

dec 23, 2016 - a load of bricks weighing 680 n a distance of 10 m across a parking lot. if if the worker pushes the cart with a constant force of 209 n, what work= force x distance = 220 n x 10 m = 2200 n.m = 2200 j i don't know where the person got it from, but the initial part, work= force x distance me.