Lösungen Physik

1

Remove["Global`*"]

R[R1_, R2_, R3_, R4_, R5_, 
        R6_] := 1/(1/R1 + 1/R2 + 1/(R3 + R4) + 1/(R5 + R6)) // 
          Together // Simplify

R[R1, R2, R3, R4, R5, R6]

R[R1_, R4_, R6_] := 
        R[R1, R1, R1, R4, R1, R6];
R[R1, R4, R6]

f[x_, y_, 
          z_, ΔR1_, ΔR4_, ΔR6_] := (Abs[D[R[R1, R4, R6], R1]] ΔR1  + 
              Abs[D[R[R1, R4, R6], R4]] ΔR4  + 
              Abs[D[R[R1, R4, R6], R6]] ΔR6) /. {R1 -> 
              x, R4 -> y, R6 -> z};
f[x, y, 
      z, ΔR1, ΔR4, ΔR6]

a

r1a = R[2, 2, 2, 10, 2, 10] // N

r1 = R[2, 10, 10] // N

b

dr1 = f[2, 10, 10, 0.05, 0.15, 0.15]

dr1u = R[2 - 0.05, 10 - 0.15, 10 - 0.15] // N

r1 - dr1

r1 - dr1u

dr1o = R[2 + 0.05, 10 + 0.15, 10 + 0.15] // N

r1 + dr1

dr1o - r1

Lineare Fehleraproximation und extreme Werte stimmen überein.

c, d

R[2, 10, R6]

D[R[2, 10, R6], R6] // Simplify

Evaluate[Simplify[D[R[2, 10, R6], R6]] == 0]

Solve[Evaluate[Simplify[D[R[2, 10, R6], R6]] == 0], {R6}] // 
      Flatten

Plot[{R[2, 10, R6]}, {R6, 0, 20}];

Plot[
        Evaluate[Simplify[D[R[2, 10, R6], R6]]], {R6, 0, 20}];

R total ist minimal für R6 = 0 und maximal für R6 = unendlich.

R[2, 10, 0]

R[2, 10, 0] // N

R[2, 10, Infinity]

R[2, 10, Infinity] // N

e

Solve[R[2, 2, 2, 10, 2, R6] == 2.8, {R6}]

Solve[R[2, 10, R6] == 2.8, {R6}]

Solve[R[2, 10, R6] == 2.8, {R6}]

Keine Lösung. R6 kann nicht negativ sein!

2

Remove["Global`*"]

m  =  2.6 10^(-3);  
    g = 9.81;   v0 = 340;
v[t_] := {Sin[45 Degree]  v0 , 
          Cos[45 Degree]  v0 -  g  t};
v[t] // N

solv2 = Solve[v[t][[2]] == 0, {t}] // Flatten

t1 = t /. solv2

t2 = 2 t1

s[t_] := {Sin[45 Degree]  v0 t, 
          Cos[45 Degree]  v0 t - 
            1/2 g  t^2};
s[t] // N

ParametricPlot[s[t], {t, 0, 25}];

ParametricPlot[s[t], {t, 0, 50}];

a

vertMax = s[t1][[2]]

b

horMax = s[t2][[1]]

Die 20 km werden längst nicht erreicht.

c

en = 1/2 m v0^2

en = 1/2  2.6 10^(-3) 340^2

3

Remove["Global`*"]

h = 1.5;  m = 1;  r = 0.02;  α = 30 Degree;  g = 9.81;  
    J = 1/2 m r^2

s = h/Sin[α]

G = m g

F = G Sin[α]

a = F/m

a

ePot = m g h

eKinRot[v_] := 1/2 m v^2 + 1/2 J (v/r)^2

eKinRot[v] == ePot

solv[3] = Solve[eKinRot[v] == ePot, {v}]

vEnd = v /. solv[3][[2]]

eRot = 1/2 J (vEnd/r)^2

eKin = 1/2 m vEnd^2

eKin + eRot == ePot

b

ωEnd = vEnd/r

ωEnd == 2  Pi/Tu

solv[31] = 
      Solve[ωEnd == 2  Pi/Tu, {Tu}] // Flatten;   Tu = 
      Tu /. solv[31]

touren/60 == 1/Tu

Solve[touren/60 == 1/Tu, {touren}]

c

vEnd == aw tE

gl1 = (s == 
            1/2 aw  tE^2); 
gl2 = (vEnd ==  aw  tE);
solv3c = Solve[{gl1, gl2}, {aw, 
            tE}]//Flatten

tE == t /. solv3c

d

Δl = 0.04;  l = 0.2;

FKin == cD  Δl

eKin  ==  1/2 cD Δl^2 // Solve

e

Remove[v]

Modellierung:

m = V1 ϱ = r1^2 Pi laenge ϱ
2m = V2 ϱ = r2^2 Pi laenge ϱ = 2 r1^2 Pi laenge ϱ 
==>  r2^2 = 2 r1^2

Jtotal = 1/2 (2m) r2^2 = m r2^2 = ΔJ + 1/2 m r1^2
ΔJ = m r2^2 - 1/2 m r1^2 = m 2 r1^2 - 1/2 m r1^2 = 3/2 m r1^2

Kontrolle mit Tabelle:
ΔJ = 1/2 m  (r2^1+r1^2) = 1/2 m (2r1^2+r1^2) = 3/2 m r1^2

JNeu  =  3  J

eKinRotNeu[v_] := 1/2 m v^2 + 1/2 JNeu (v/r)^2

eKinRotNeu[v] == ePot

solv[3] = Solve[eKinRotNeu[v] == ePot, {v}]

vEndNeu = v /. solv[3][[2]]

eRotNeu = 1/2 JNeu (vEndNeu/r)^2

4

Remove["Global`*"]

d = 0.05;  s1 =  0.1;  
    temp1 = 273.15 + 18;   temp2 = 273.15 + 100;  
    p1 = 1;
V[s_] := d^2 Pi / 4 s;
V[s1]

a

solv41 = Solve[p1/temp1 == p2/temp2, {p2}] // Flatten

p2 = p2 /. solv41

b

solv42 = 
      Solve[p2 V[s1]  == p1 V[s2] , {s2}] // Flatten

s2 = s2 /. solv42

5

Remove["Global`*"]

m = 2;  M = 5;  s = 2;  g = 9.81;

a

solv51 = Solve[a (m + M) == m g, {a}] // Flatten

a = a /. solv51

solv52 = Solve[1/2 a t^2  ==  s, {t}] // Flatten

tEnd = t /. solv52[[2]]

b

vEnd = a tEnd

eKinm = 1/2 m vEnd^2

Solve[m g s  == (m + M)/2 v^2, {v}]

c

solv53 = 
      Solve[{1/2  M vEnd^2 == 1/2  M v1^2 + 1/2  M v2^2,  
            M vEnd == M v1 + M v2}, {v1, v2}] // Chop

d

eKin2 = 1/2 M v2^2  /.   solv53[[1]][[2]]

6

a

Remove["Global`*"]

m = 1;  h = 1;  s = 0.02;  g = 9.81;

ePot =  m g h;
gl1 = (eKin == ePot);
gl2 = (eKin == 1/2 m v^2);
ePot

solv61 = Solve[{gl1, gl2}, {v}] // Flatten

vEnd = v /. solv61[[2]]

a = vEnd/t

s == 1/2 a t^2

solv62 = Solve[s == 1/2 a t^2, {t}] // Flatten

tVer = t /. solv62

imp =  m vEnd

F = imp/tVer

b

Remove["Global`*"]

u = 230  volt; temp2 = 100 K; temp1 = 20 K; V = 0.1^3 m^3;t = 30 sec;ϱ = 1 kg  /(0.1^3 m^3);spW=4187 J/(kg  K);

W[i_] =  u i t  /. {sec volt -> J/amp}

Solve[
        W[i] ==  spW (temp2 - temp1) V ϱ , {i}]

c

Remove["Global`*"]

k = 6.674 10^-11;   mE =  5.9736  10^24 ; m = 892;  
    tU = 6.4 60 60;  rErde = 6371 10^3

v = 2 r Pi/tU;  Fz[r_] := m v^2/r;  Fz[r]

F1[r_] := k m mE/r^2;  F1[r]

solv6c = Solve[F1[r] == Fz[r], {r}] // Flatten

rU = r /. solv6c[[3]]

rU - rErde

Anderer Weg

Remove["Global`*"]

k = 6.674 10^-11 ;  mE =  
        5.9736  10^24 ;m = 892 ; tU = 
        6.4 60 60 ; rErde = 
        6371 10^3 ;

gl1 = (m r ω^2  ==  γ M m / 
                  r^2  /.  {γ M / r^2  ->  
                gr,   ω -> (2  Pi)/T}) // 
        Simplify

solve6c1 = 
      Solve[{γ M == gr r^2 /. {γ M -> g rErde^2}}, {gr}] // 
        Flatten

gr = gr /. solve6c1

solv6c2 = (Solve[gl1, {r}] /.  {g -> 9.81, T -> tU}) // 
        Flatten

r = r /. solv6c2[[3]];   r - rErde

d

Remove["Global`*"]

s = 0.25;  λ = 2 s; v = 337;

Solve[v == λ f, {f}]

Das ist etwa e gegen f

e

Remove["Global`*"]

ϑ1 = 20.5;  ϑ2 = 
      2.6;  Δϑ1 = 
      0.3;  Δϑ2 = 0.3;   t1 = 
      18;  Δt = 1;

f[
          t_, ϑ1_, ϑ2_] := ϑ1 + ϑ2 Sin[2  Pi t/24];
Plot[{f[t, 20.5, 2.5], 20.5}, {t, 0, 48}];

Plot[{f[t, 20.5, 5], f[t, 20.7, 5], f[t, 20.3, 5]}, {t, 17, 
          19}];

Plot[{f[t, 20.5, 5], f[t, 20.7, 5], f[t, 20.3, 5]}, {t, 16, 
          20}];

Plot[{f[t, 20.5, 2.5], f[t, 20.5, 2.3], f[t, 20.5, 2.7]}, {t, 16, 
          20}];

M1 = {f[t, 20.5, 2.6], f[t, 20.2, 2.3], f[t, 20.8, 2.3], 
          f[t, 20.2, 2.9], 
          f[t, 20.8, 2.9]}  /.  {t -> 18 }

M2 = {f[t, 20.5, 2.6], f[t, 20.2, 2.3], f[t, 20.8, 2.3], 
          f[t, 20.2, 2.9], 
          f[t, 20.8, 2.9]}   /.  {t -> 17 }

M3 = {f[t, 20.5, 2.6], f[t, 20.2, 2.3], f[t, 20.8, 2.3], 
          f[t, 20.2, 2.9], 
          f[t, 20.8, 2.9]}   /.  {t -> 19}

m1 = Min[Union[M1, M2, M3]]

m2 = Max[Union[M1, M2, M3]]

(Max[Union[M1, M2, M3]] - Min[Union[M1, M2, M3]])/2

f1[t_, h1_, 
          h2_, Δt_, Δh1_, Δh2_] := (Abs[Evaluate[D[x + y Sin[2  Pi z/24], 
                      x]]] Δh1  + 
Abs[
                  Evaluate[
                    D[x + y Sin[2  Pi z/24], 
                      y]]] Δh2  + 
Abs[
                  Evaluate[
                    D[x + y Sin[2  Pi z/24], 
                      z]]] Δt ) /.  {x -> h1, 
            y -> h2, z -> t};

diff = f1[18, 20.5, 2.6, 2, 0.1, 0.2]

val = f[18, 20.5, 2.6]

m3 = val - diff

m4 = val + diff

{m1, m3, val, m4, m2}

f

Remove["Global`*"]

v1 = 0;  h1 = 1;  h2 = 0;  g = 9.81;

gl = ( ϱ v1^2/2  +  ϱ g h1  +  
              p ==  ϱ v2^2/2  +  ϱ g h2  +  
              p);
solv6f = Solve[gl, {v2}]

v2 = v2 /. solv6f[[2]]

Anderer Weg:

Sqrt[2 g h1]

Oben muss soviel Wasser reinkommen wie unten rausgeht. Da mit den Rohrdurchmessern auch die Volumen pro Zeit, also die Geschwindikgeiten gleich sind, ist oben v =4.429...