Esercizi svolti d'esame - Fisica 1

Esercizio 1.12

h: 126m 18m

g: 9.81 m/s²

x = v₀t - 1 _{t = v0/q}

x = v₀t - 1/2 q t² => x = (v₀²)/g - 1/2 g t²

x - v₀²/_g = 0 => t = v₀/g

x = v₀( v/q ^{y - q}

x - q = 0 => -1/2 q t² + h = 0, 1/2 q t² = h; t = √(2h/q), x = s

h x/q = s

v_x = v₀ = 200 m/s

v_y = q -(4.81

_len

0)

y_j = v_yi t

x = s_t

v_y = h + v_yt = 1/2gt²

mi ricavo v_y y = -0

v_y = 1/2gt² h = 9.81 m/s² 8s² 400m = 86.1 m/s

2)

e v_x = v sin θ = 18.6 m/s

s = u_x x t = 149.1 m

3)

v_x = u_x = 18.6 m/s

v_y2 = u_yj - gt = - (10.7 m/s, 9.81 m/s² 88s) = 89.2 m/s

Esercizio 1.1

m₁ = 4.0 kgm₂ = 2.0 kgF = 6 NF = m · a

F + F_1,2 - F_2,1 = (m₁ + m₂) · acon F_1,2 = F_2,1

=> F = (m₁ + m₂) · a => a = F / (m₁ + m₂) = 6 N / 6.0 kg = 1 m/s²

F_1,2 = m₁ · a = 4.0 N => F_2,1 = 4.0 N

Per il 1^o blocco:

F + F_1,2 = m₁ · a => F_1,2 = m₁ · a - F = -2.0 N

F + F_2,1 = m₂ · a => F_1,2 = m₂ · a - F = -4.0 N

Esercizio 2.2

m = 20 g = 0.020 kgU₀ = 10 m/sθ = 25°

1) F = m · aF

p

• F_p = F_attrito = m · g · sin θ
• F_p · cos θ = Nt · g · sin θ = a

t = g · sin θs = 12

2) v = v₀ + a · t = at² / 2s = v₀ · t - 1/2 · a · t²t₁s · t = 12 m

3) v_y = v₀y - g · t = (v₀cos(θ)²) - 1/2 · g · t²

y = 12 sin(25°) = 5.0 m

Esercizio 28

M₁ = 1,65 Kg

M₂ = 3,3 Kg

β = 30°

μ_d = 0,226

μ_d = 0,113

F = ma

N₁ - Mgcosβ + N₂ - M₂gcosβ = 0

per il corpo 1:

• y: N₁ = Mgcosβ
• x: T - μ_d1Mgcosβ +μ₁gsinβ = M₁a

per il corpo 2:

• N₂ = Mgcosβ
• T + μ_d2M₂gcosβ + T₂gsinβ = M₂a

M₁a + μ_d1M₁gcosβ - M₁gsinβ = M₁(a + μ_d1gcosβ - gsinβ)

T = M₂gsinβ - μ_d2M₂gcosβ - M₂a

Ma + M₁a = M₁gsinβ + M₁gsinβ - M₂M₂gcosβ - M₁M₁gcosβ

4 esame 2016

T_eq = -3°C

T₁ = 20°C

T₃ = 10°C

c_GH = 3.3 · 10⁵ J/Kg

c_GH = 2051 J/Kg K

c_H2O = 4186 J/Kg K

• Calore scambiato dal ghiaccio: Q_GH = c_GH m₁ (T_eq - T₁)
• Calore scambiato da H₂O: Q_H2O = c_H2O m₃ (T_fus - t₃) + λ_GH m₃ + c_GH m₃ (T_eq - T_fus)
• Calore scambiato dal corpo: Q_corpo = c_m (T_eq - T₂)

=> m₁ c_GH (T_eq - T₁) + c_H2Om₃ (T_fus - T₃) - λ_GH m₃ + c_GH m₃ (T_eq - T_fus) + c_m (T_eq - T₁) = 0

m₁ = ( -c_H2O m₃ (T_fus - T₃) + λ_GH m₃ - c_GH m₃ (T_eq - T_fus) - e_m m₁ (T_eq - T₂) ) / e_GH (T_eq - T₁)

m₁ = -4186 (0 - 10) + (3.3 · 10⁵)·0.8 - (2051·0.8) [-3 - 0]) / (380·0.4 [-3 - 60]

m₁ = 9.1 Kg

n.4 esame 2014

m_H2O = 200 gt_i = 30 °Ct_p = t₂ = 10 °Cλ_GH = 3.3 ⋅ 10⁵ J/gc_H2O = 4186 J/(kgK)

Q_GH + Q_H2O = 0

m ⋅ (λ_GH + c_H2O (t₂)) = - c_H2O m_H2O (t₂-t_i)

m = - c_H2O m_H2O (t₂-t_i) / (λ_GH + c_H2O t₂)

= (-4186 ⋅ 0.2 ⋅ (-20)) / (3.3 ⋅ 10⁵ + 4186 ⋅ 10)

= 0.045 kg = 45 g

n.5 esame 2014

1 mol (gas monoatomico)

C_V = 3/2 R

C_P = 5/2 R

p_A = 10⁵ Pa

V_A = 22.4 l = 0.224 m³

T_B = 100°C = 373 K

A -> B: isobara

Q_AB = n C_P (T_B - T_A) = 5/2 R (373 - 293) = 2.16 ⋅ 10³ = 1.66 ⋅ 10³ J Q_AB

Q_BC = nRT_B ln (V_A / V_B) = nRT_B ln (V_A / nRT_B) = -1.00 ⋅ 10³ Q_BC

Q_CA = n C_V (T_A - T_B) = 3/2 R (293 - 373) = 3/2 R (293 - 373)

T_A = p_A V_A / nR = 3.16 m³

Q_tot = Q_AE + Q_BC + Q_CA = 1.66 - 1.00 + 0.998 ⋅ 10³

69

0.73 J

= 3.38 ⋅ 10³ J

V_A = V_B = 2V_C

T_A = 3T_C

monatomico (C_p = ⁵/₂ R; C_v = ³/₂ R)

A → B isocòra

W = 0

Q = nC_V(T_B − T_A) → T_B < T_A ⇒ Q_AB ceduto < 0

B → C isobara

W_BC = nR(T_C − T_B)

= nR(T_C − 2T_C) = nR(−T_C)

Q < 0 ceduto

V_C = ^T_C/_TB

→ V_C = ^T_C/_2VC → ^V_C/_2VC

C → A :

ΔU = Q_CA − W_CA = nC_vΔT =

= nC_v(T_A − 3T_C) =

= nC_v(3T_C − T_C)

W_CA = ^(V_B−V_C)/₂(P_A+P_C) = ^(2V_C−V_C)/₂(P_A+P_C) =

= ^V_C/₂(P_A+P_C)

p_C = ^nRT_C/_VC → T_B = ^nRT_C/_VC − ^2V_C/_VC =

= ^3T_C/₂T_C = 2T_C

p_A = ^nR3T_C/_2VC

P_B = P_C = ^nRT_B/_VB = ^nRT_B/_2VC

^V_C/_5VC

nRT_C

W_B = −nRT_C

W_CA = V_C(³/₂nRT_C + nRT_C)¹/₂ =

= (³/₂nRT_C + nRT_C)¹/₂ = ³/₄nRT_C +¹/₂nRT_C

⁵/₄nRT_C

η = ¹/₁₇ = 0,06 → 6%