Question

cum se face problema astaa? puteti sa ma ajutati va roog? mersiii anticipat

Answer 1

1.
a.
Δt=3s
v=340m/s
Δt'=2s
d=?
d=v×Δt
v=340×3=1020
v=1020m=1,02km
b.
Δt'=2s
d'=v×Δt'
d'=340×2=680m
d'<d ⇒furtuna se apropie.


2.
v1=90km/h=90/3,6=25m/s
t=1/50s
d=60cm=0,6m
a.
v=d/t
v=0,6/(1/50)=0,6×50=30
v=30m/s
v>v1 ⇒a depasit viteza legala
b.
Viteza medie se afla impartind distanta totala, d1, la timpul de parcurgere, t1.
Mobilul parcurge distanta d/2 cu viteza v1, in timpul t1'
d/2=v1×t1'⇒t1'=d/2v1
El parcurge restul din distanta( adca d/2) cu viteza v2=20m/s, in timpul t1''
d/2=v2×t1''⇒t1''=d/2v2
Atunci viteza medie este:
vm=(d/2+d/2)/(t1'+t1'')
vm=d/(d/2v1+d/2v2)=d/d(1/2v1+1/2v2)
Simplificam cu d si, in paranteza de la numitor, aducem la acelasi numitor
vm=1/[(v2+v1)/2v1v2]=2v1v2/(v1+v2)
vm=2×25×20/(20+25)=1000/45=22,23m/s
c.
Procedam simlar ca la punctul b.
t3'=d/(2×4v)=d/8v
t3''=d/(4×2v)=d/8v
t3'''=d/4v
Atunci:
vm=d/(d/8v+d/8v+d/4v)
vm=1/[(1+1+2)/8v]=1/(4/8v)=1/(1/2v)=2v
vm=2v