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# Practica uno y dos

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Practica 1 y 2 de Fisica 1

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### Practica uno y dos

1. 1. TSU. García José R.C.I.: 19.407.757Guarenas, Mayo del 2013
2. 2. xa= ½(g.t ²)xa= ½(9,8.0,035 ²)xa= ½(0,012)xa= 0,006 mtsxb= ½(g.t ²)xb= ½(9,8.0,007 ²)xb= ½(0,0004)xb= 0,0002 mtsdT= ha+hbdT= 0,006mts+0,0002 mtsdT= 0,0062 mtsPractica Nº 1
3. 3. xa= ½(g.t ²)xa= ½(9,8.0,090 ²)xa= ½(0,079)xa= 0,039 mtsxb= ½(g.t ²)xb= ½(9,8.0,021 ²)xb= ½(0,0043)xb= 0,0021 mtsdT= ha+hbdT= 0,039mts+0,0021 mtsdT= 0,041 mtsPractica Nº 1
4. 4. Vx= cos θVx= 0,98Vy= sen θ-g.tVy= 0,17-9,8.0,037Vy= 0,35VT= Vx+VyVT= 0,98+0,35VT= 0,63 mts/shmax= V ².sen θ/2.ghmax= (0,63)².0,17/2.9,8hmax= 0,066/19,6hmax= 0,003 mtstv= 2.V.sen θ/gtv=2.0,63.0,17/9,8tv= 0,21/9,8tv= 0,021 sAlcance horizontal:R= 1/g.(V².sen2. θ)R= 1/9,8(0,63².0,34) = 1/1,29 = R=0,75 mtsPractica Nº 2
5. 5. Vx= cos θVx= 0,86Vy= sen θ-g.tVy= 0,54-9,8.0,054Vy= 0,08VT= Vx+VyVT= 0,86+0,08VT= 0,78 mts/shmax=- Vy²/2.ghmax= -(-0,08) ²/19,6hmax= 0,003 mtstv= 2.V.sen θ/gtv= 2.0,78.0,54/9,8tv= 0,085 sAlcance horizontal:R= 1/g.(V².sen2. θ)R= 1/9,8(0,78².0,34) = 1/9,8.0,20 = R=0,51 mtsPractica Nº 2
6. 6. Vx= cos θVx= 0,70Vy= sen θ-g.tVy= 0,707-0,25Vy= 0,45VT= Vx+VyVT= 0,70+0,45VT= 0,25 mts/shmax=- Vy²/2.ghmax= -(0,45) ²/19,6hmax= 0,010 mtstv= 2.V.sen θ/gtv= 2.0,45.0,70/9,8tv= 0,063/9,8tv= 0,064 sAlcance horizontal:R= 1/g.(V².sen2. θ)R= 1/9,8(0,25².1) = 1/9,8.0,62 = R=6,07 mtsPractica Nº 2
7. 7. Vx= cos θVx= 0,5Vy= sen θ-g.tVy= 0,86-9,8.0,03Vy= 0,57VT= Vx+VyVT= 0,5+0,57VT= 1,07 mts/shmax=- Vy²/2.ghmax= -(0,57) ²/19,6hmax= 0,016 mtstv= 2.V.sen θ/gtv= 2.1,07.0,86/9,8tv= 1,84/9,8tv= 0,18 sAlcance horizontal:R= 1/g.(V².sen2. θ)R= 1/9,8(1,07².0,86) = 1/9,8.0,89 = R=0,10 mtsPractica Nº 2
8. 8. Vx= cos θVx= 0,17Vy= sen θ-g.tVy= 0,98-9,8.0,027Vy= 0,72VT= Vx+VyVT= 0,17+0,72VT= 0,89 mts/shmax=- Vy²/2.ghmax= -(0,72) ²/19,6hmax= 0,02 mtstv= 0,17 stv= 2.0,89.0,98/9,8tv= 1,74/9,8tv= 2.V.sen θ/gAlcance horizontal:R= 1/g.(V².sen2. θ)R= 1/9,8(0,89 ².1,96) = 1/9,8.1,54 = R=0,066 mtsPractica Nº 2