4º ESO: SOLUCIONES AL EXAMEN PARCIAL DEL 07.11.11

1.-

a) 36 km/h x 1000m/1km x 1h/3600s = 10 m/s

b) 30 km/min x 1000m/1km x 1min/60s = 5000 m/s

2.- v1 = 1200 cm/s    t1 = 9s

      v2 = 480 cm/s      t2 = 7s

a)      V= Δs/t à Δs = v·t

Δs1 = 1200 cm/s · 9s = 10800 cm

Δs2 = 480 cm/s · 7s = 33600 cm

Δs = Δs1 + Δs2 = 14160 cm

b)      V= 14160cm / 16s = 185 cm/s

3.- v = 80 km/h · 1000m/1km · 1h/3600s = 22, 22 m/s

90 min · 60s/1min = 5400 s

S=So + Vt

S1 = 22,22m/s · (5400s + t)

S2 = 27,78m/s · t

S1=S2 à 27,78m/s · t = 22,22m (5400s + t) à t = 23715,42s

Sustituyendo: s = 27,78m/s · 23715,42s = 658814,37m

4.- a1 = 1m/s²    t=15s     Vo=0m/s

a2 = -25 cm/ s² ·10^-2 = -0,25 m/s²

a3 = ?    t=5s       Vf=0m/s

1)      S=So + Vt +1/2at²t

S = 0 + 0t + ½·1m/s²·(5s)² à S = 112,5m

V = Vo + at = 0 + 1m/ s² · 15s à Vmax = 15m/s

2)      S = 112,5m + 15m/s · 10s -1/2(-0,25m/ s²)·(10s)²=275m

V = 15m/s + (-0,25m/s²)·10s = 12,5m/s

3)      0 = 12,5m/s +a · 5s à a₃ = -12,5m/s : 5s = -2,5m/s²

S = 275m +12,5m/s · 5s + ½·(-2,5m/s²)(5s)² = 306,25m

5.-

r = 6380 km

T=24h=86400s

ω = Φ/T = 2π/86400s = 7,27 · 10^-5 rad/s

ac = v²/r = (ω·r)² /r= ω²·r= (7,27 · 10^-5 rad/s) ²·6380·10^3m = 54398,62 m/s²