ejercicios fisica

universidAd militAr nuevA grAnAdA
TAller de REFUERZO
SAMANTA LEON GOMEZ


5800071





1.

SOLUCION
a)
ac
1mi = 5280 ft
= 88 ft / s
ac = w2/r = 25ft = 25ft = 309,76 ft/s2ac = 309,76 ft/s2
b)
ac
1mi = 1 609,344 m
1ft=0,3048m

= 26,82m / s
ac = w2/r = (25*0,3048m) = (7,62m) = 94,4 m/s2
ac = 94,4 m/s2


2.

a)
(1 libra = 4.458 newton)
(1 libra=.454kg) = 2400 libras equivale a 1090 kilogramos
ac= (a^2/r)= (13.4m/s)(13.4)/61m = 2.9436 m/s
F=(1090KG)X(2.9436M/S cuadrados)=30207.34 newtons
F = 30207.34 newtons / 4.458 newton
F =720libras
b)
V=(g.r.Ue)^1/2 
13.4=(9.8.63.Ue)^1/2 
(13.4)^2=5978.Ue 
176.5/5978=Ue 
0.3= Ue 






3.

A)
tan-1(v^2/rg)9
tan-1(2.44)9
tan-1(21.96)
0.99

B)
Fu=(v^2/rg)Fu=37Mi/h^2/150ft*9.8
Fu=3600Mi/h/1470
Fu=2.44








4.


m = 53 g = 0,053 Kg
Cuerda = 1,4 m
Radio = 25 cm
L= longitud de la cuerda




a) Calcular la velocidad del guijarro: V=?

W= m.g
W = 0,053 Kg . 9,8 m/s2
1N= 1kg.m/s2
W= 0,5194kg.m/s2 = 0,5194 N
Senα= r/L = 0,25 m / 1,4m = 0,17
sen 2α + cos2 α = 1
α = sen-1(0,17) = 9,78o
cos2 α = 1 - sen 2α = 1 – (0,17)2 = 1-0.0289 = 0,9711
cos α = = 0,985
tan α = sen α/cos α = 0,9711/0,985= =0,172

tan α= Fc /w
Fc= tan α . w = 0,172 x 0,5194 kg.m/s2 = 0,0893N
Fc= mw2/r
w2=Fc.r/m= 0,0893 kg.m/s2 X 0,25 m / 0,053 Kg= 0.42122642/s2
w = = 0.64901958/s
v= w.r
v= 0.64901958/s x 0,25 m = 0,1622 m/s

b) a=?
a= v2/ r = (0,1622 m/s)2/ 0.25 m
a= v2/ r = (0.02630884 m2/s2)/ 0.25 m = 0.10523536 m/s
c)T= ?
cos α = w/ T

T = w/ cos α = 0,5194 N / 0,985 = 0.527309645 N


5.

a)


v= 1,24X10^-13/5.3x10^-11 = 0.23x10^-24
b)
= (6.6x10^15rev/ s)/ 5.3x10^-11 =1,24X10^-13

c)

F= (34.98X10^4)*(1,24X10^4)
F= 43.37x10^4






6.


2da. Ley del Newton al disco:
ma = T de donde
0 = N- mg
A la pesa que está en reposo (a = 0) : 0 = T- Mg  T =...