Fisica general. SOLUCIONES EXAMEN PARCIAL A DISTANCIA
Enviado por mgsilveyro • 16 de Mayo de 2021 • Práctica o problema • 6.174 Palabras (25 Páginas) • 121 Visitas
SOLUCIONES EXAMEN PARCIAL A DISTANCIA
1. -[pic 1]
L = 36 CM 10 KG 20 KG 10 KG 36 CM 20 KG
La . F = Lb . F (360 CM – A) . 10 KG = 20 KG . A
360 CM . KG – 10 KG . A = 20 KG
A = 36 – B 360 CM . KG = 20 KG . A + 10 KG . A
B = 36 – A 360 CM . KG = 20 KG . A + 10 KG . A
360 CM . KG = (20 KG + 10 KG) . A
B = 36 CM – A 360 CM . KG = 30 KG . A[pic 2]
B = 36 CM – 12 CM 360 CM . KG = A 12 CM = A [pic 3]
B = 24 CM 30 KG
2. -
20 CM 60 CM 10 CM[pic 4][pic 5][pic 6][pic 7][pic 8][pic 9]
6 KG 6 KG 7 KG 3 KG 8 KG
L = 100 CM ΣF = - 6 KG + 6 KG – 7 KG – 3 KG + 8 KG
F1 = - 6 KG ΣF = - 16 KG + 14 KG
F2 = 6 KG ΣF = - 2 KG
F3 = - 7 KG
F4 = - 3 KG MF1 = - 6 KG . 0 CM = 0 KG . CM
F5 = 8 KG MF2 = 6 KG . 20 CM = 120 KG . CM
MF3 = - 7 KG . 80 CM = 560 KG . CM
MF4 = - 3 KG . 90 CM = 270 KG . CM
MF5 = 8 KG . 100 CM = 800 KG . CM
90 KG . CM = 2 KG . D[pic 10]
90 KG . CM = D 45 CM = D[pic 11]
2 KG
3. -
0 KM 1400 KM[pic 12]
V1 = 700 KM/H
V2 = 80 KM/H
[pic 13]
V = D T = D T = 1400 KM T = 1400 KM T = 1,79 H[pic 14]
T V 700 KM/H + 80 KM/H 780 KM/H
4. - [pic 15]
T ?
0 M/S 24 M/S
A = 8 M/S2 D [pic 16]
A = VF – VI A . T = VF – VI T = VF – VI T = 24 M/S – 0 M/S
T A 8 M/S2
[pic 17][pic 18]
T = 24 M/S D = VI . T ± 1 A . T2 D = 0 M/S . 3 S + ( 1 . 8 M/S2 ) . (3 S)2 [pic 19][pic 20]
8 M/S2 2 2[pic 21][pic 22]
T = 3 S D = 8 M/S2 . 9 S2 D = 4 M/S2 . 9 M/S2 D = 36 M
2
5. - V = 0,5 M/S H = 0,78 M G = 9,8 M/S2
H = G . T2 T = √ H . 2 D = V . T
2 G[pic 23][pic 24]
T = √ 0,78 M . 2 D = 0,5 M/S . 0,39S
9,8 M/S2 D = 0,195 M[pic 25][pic 26]
T = √ 1,56 M
9,8 M/S2[pic 27][pic 28][pic 29]
T = √ 0,15 S2
T = 0,39 S
6. -
Vi = 20 M/S H = G . T 2 H = (Vf – Vi) 2
2 2.G
T = √ H . 2 H = 400 M2/S2
G 2 . 9,8 M/S2[pic 30][pic 31]
T = √ 20,40 M . 2 H = 400 M2/S2[pic 32]
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