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najat
Answers (23)
F1/A1=F2/A2 F1=F2A1/A2 F1=1600N*3.75cm2/75CM2= 80N
F1= F2*A1/A2 = 1000N*0.5/2= 250N
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
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
T=NBAI I=T/NBA I=0.6/30*0.31*(0.18)^2 I=1.99A
t=x/c t=3.2*10^-3/ 3*10^8 = 1.06*10-11 sec
v=(2hg)^1/2 v=(2*93*980)^1/2 v=426.9cm=4.269m
The angle of incidence equals the angle of reflection = 10º
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
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
m=I/o = 16mm/8mm=2 m=hi/ho hi-m*ho=2*4mm=8mm
f=0.005N
2476 Pa /m
t=s/v=5cm.s/30cm=0.16s
565.3W
458.5cal
LfA= 14411.7J/kg, LfB= 32352.9J/kg, QA= 98„e103J
31760J=31.76KJ
75807pa
1.399m
115807m
14945.98Pa
a. 0.17 m/s
