najat

This page lists questions and answers that were posted by visitors named najat.

Questions

The following questions were asked by visitors named najat.

Answers

The following answers were posted by visitors named najat.

LfA= 14411.7J/kg, LfB= 32352.9J/kg, QA= 98„e103J

t=s/v=5cm.s/30cm=0.16s

m=I/o = 16mm/8mm=2 m=hi/ho hi-m*ho=2*4mm=8mm

when P=1200W , R=12Ù the current is I^2=P/R=100 , I=10A So the power P=? R=6Ù I=10A P=I^2R= 100*6=600W

shorten length= F.l/A.E = 3.2*10^4*16*10^2/4.4*10^-4*1.5*10^10 =7.75*10^-4 m = 0.775mm

The angle of incidence equals the angle of reflection = 10º

v=(2hg)^1/2 v=(2*93*980)^1/2 v=426.9cm=4.269m

t=x/c t=3.2*10^-3/ 3*10^8 = 1.06*10-11 sec

T=NBAI I=T/NBA I=0.6/30*0.31*(0.18)^2 I=1.99A

The surface tension of soapy water is 25*10^3 N/m PG = 4T/r PG=4*25*10^3N/m/0.0002m PG= 500Pa

Q=vA Q=5L/min=8.3e-5 m^3/sec A= pi*r^2=3.14*4*10^-4=0.001256 m2 v=Q/A v=8.3e-5/0.001256 = 0.006m/sec = 0.6cm/sec

F1= F2*A1/A2 = 1000N*0.5/2= 250N

F1/A1=F2/A2 F1=F2A1/A2 F1=1600N*3.75cm2/75CM2= 80N