Read correction series 2 interaction mecanique text version

2

3 S ) R=8N P=mg P=12N F=8,5N 2 1

/ R

P F

j

(

=45°

i

1cm 4N

(

)

3 (Ox,Oy) P P = Px + Py

sin = Px P Py P

P = Px i + Py j cos =

Px = P sin Py = P cos

Allal Mahdade

http://sciencephysique.ifrance.com

1

2 S1 : S1 1 P1 : S1 1

R1 :

f 1/ S1 : S1 f 2 / S 1 S1

S2

2 2 P2 : S2

R2 :

f 2 / S 2 S2 .

R

(S1,S2) 2

f 1/ S1

2 3 . P : (S1 ,S2) 1 f 2 / S S1 1 2 3

f 2 / S 2 S2

f 2 / S 2 f 2 / S1 : f 2 / S1 + f 2 / S 2 = 0 .

0,5bar

:

20bar 0 14

14 20 5 10 Pa 1bar

P = 1bar + 14 × 0,5bar P = 8bar = 8.105 Pa

4 F

P=

1 2

F F = P ×S S S =R 2 F = P × R 2

: R=2.10-2m P=0,5.105Pa F=63N

Allal Mahdade

http://sciencephysique.ifrance.com

2

h

h

.

5 1 p - p 0 = gh p = p 0 + gh h p p0 . h=60m 2

p = p 0 + gh 10-3 kg p 0 = 10 Pa = -6 3 = 103 kg / m 3 g=10N/kg h=60m 10 m 5 3 p=(10 + 10 × 10 × 60)Pa

5

p = 7 × 105 Pa

F p = F = p ×S S

2

3

d d S = F = p × 2 2 5 F=5,5.10 N :

2

Allal Mahdade

http://sciencephysique.ifrance.com

3

Information

correction series 2 interaction mecanique

3 pages

Find more like this

Report File (DMCA)

Our content is added by our users. We aim to remove reported files within 1 working day. Please use this link to notify us:

Report this file as copyright or inappropriate

108982