=====Physics 2 (Essay)===== (1a) F= Force, m= Mass, a= Acceleration F is directly proportional to (m) F/a=ma/a m=F/a (N/m) (1b) Dimensionally analysis for m=M since F= MLT -2 a=LT -2 hence, F/a=MLT -2/LT -2 :. Dimension for m = Dimension for F/a which is M this implies that the equation is dimentionally correct (2) U=20m/s S=125m t=? g=10m/s -2 S=ut + i/2at 2 125= 20t + 1/2(10)t 2 125=20t + 5t 2 5t 2 + 20t-125=0 Dividing through by 5 5t 2/5 + 20t/5 -125/5 =0/5 t 2 + 4t – 25=0 solving the equation quadratically, using general method t= -bt + sqrt(b 2)-4ac/2a t= -bt – sqrt(b 2)-4ac/2a a=1, b=4, c= -25 t= -4 + sqrt(4 2-4(1)(-125)/ 2*1 t= 4 + sqrt(16+500)/2 t= 4+ sqrt(416)/2 t= 4 + 20.4/2 or 4-20.4/2 t= 24.4/2 or -16.4/2 hence, t=12.25 or -8.25 so the time it takes the ball to reach the ground (t)=12.25 (6) m= 9.1*10 -31kg v=6.6*10 7m/s wavelenght= ? h=6.6*10 -34Js wavelenght =h/meV V= sqrt(2eV/me) wavelenght= 6.6*10 -34/ (9.1*10 -34)* (6.6*10 -7) =1/9*10 7 wavelenght= 0.11*10 7 wavelenght=(1.1 *10 6)1electrovolt 1electrovolt = 1.6*10 -19 wavelenght=(1.1 *10 -6) (1.6*10 -19) =1.76*10 -13m (7) l1 =5.0*10^-2m=e1 l2=5.5*10^-2m=e 2 l3= 6.5*10^-2m=e3 load(m1)=0.2N m=ke k=m/e 0.2/(5.5*10^-2 )-(5*10^-2) = m2/ (6.5*10^-2)- (5.5*10^-2) 0.2/0.5=m2/1.0 0.5m2/0.5=0.2/ 0.5 m2= 0.4N The additional load= 0.4N Q8b We may say that an object at rest is in equilibrium or in static equilibrium. An object at rest is described by Newton’s First Law of Motion. An object in static equilibrium has zero net force acting upon it. The First Condition of Equilibrium is that the vector sum of all the forces acting on a body vanishes. This can be written as F = F 1+ F 2+ F3 + F 4+. . . = 0 HURRY INVITE YOUR FRIENDS EVERYTHING HAS BEEN DROP THERE AT THE BELOW LINK ALL U HAVE TOO DO CLICK THE LIKE AND JOIN TOO GET BOTH THE FULL OBJ AND THEORY LINK https://m.facebook/groups/894592387237388
Posted on: Tue, 16 Sep 2014 10:06:14 +0000
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