Físixca

ALUMNO: TRINIDAD FLORES ROQUE MARTIN
SECCIÓN: “C”
1

T=2πR+KxR g12→T=2πR+KxR g12
T=LXL L12 T-1
TL0=L(x-32)T0=x-32

∴ x=32

2

a=(k5k7k9)(k3k6) Dato: F=K0V1+K1V22+K2V33+…F=K3V44→MLT-2=K3L4T-4→K3=ML-3T2
F=K5V66→MLT-2=K5L6T-6→K5=ML-5T4
F=K6V77→MLT-2=K6L7T-7→K6=ML-6T5
F=K7V88→MLT-2=K7L8T-8→K7=ML-7T6F=K9V1010→MLT-2=K9L10T-10→K9=ML-9T8
a=(ML-5T4ML-7T6ML-9T8)(ML-3T2ML-6T5)→a=ML-21T18ML-9T7→a=ML-12T11

∴a=K12

3Q=3,16V98,49D+V0 Dato: Q=lbpie-s

98,49D=V0→98,49D=V0→98,49ML-3=LT-1
98,49=M-1L4T-1 → a S.I.→98,49(0,3048)4m40,4536Kg-S=1.8740m4Kg-S
Q=3,16V98,49D+V0→Q=3,16LT-1LT-1

Reemplazamos el dato:
ML-1T-1=3,16LT-1LT-13,16=ML-1T-1 →a S.I. → 3,160,4536Kg0,3048m-s=2,6445Kgm-s
V=PiesS →a S.I. → V= ms →V(0,3048)ms
D=lbPies3 →a S.I. → D= Kgm3→D(0,4536)(0,3048)3Kgm3 →D(16,018)Kgm3
V0= PiesS →a S.I. → V0= ms →V0(0,3048)ms
Reemplazando en la ecuación: En S.I.Q=2,64450,3048V1,874016,018D+0,3048V0

∴ Q=0,806 V30,019 D+0,3048 V0

4

I=FT…α I=WZ+MZ…β
Deα:
I=MLT-2T→I=MLT-1
De β:
I=ML2T-2Z+MZ
Reemplazamos α en β :
MLT-1=ML2T-2Z+MZ→ML2T-2Z=MZ

∴Z=LT-1