Side 1 af 2 Fysik B eksamen december 2008 â Løsningsforslag ...
Side 1 af 2 Fysik B eksamen december 2008 â Løsningsforslag ...
Side 1 af 2 Fysik B eksamen december 2008 â Løsningsforslag ...
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<strong>Fysik</strong> B <strong>eksamen</strong> <strong>december</strong> <strong>2008</strong> – Løsningsforslag:<br />
Opgave 1<br />
a) Q rør, fortætning = Q f, vand = m f, vand · L f, vand = 18,4 kg · 2260 kJ/kg = 41,6 MJ<br />
6<br />
E<br />
b) P =<br />
Δ t<br />
= 41,6 ⋅10<br />
J<br />
= 481 W<br />
2<br />
24 ⋅ 60 s<br />
Opgave 2<br />
a) E foton = h · f = 6,63·10 -34 J·s · 4,6·10 14 s -1 = 3,05·10 -19 J<br />
8<br />
3,00 ⋅10<br />
m/s<br />
b) λ foton = v foton / f foton = = 6,5·10 -7 m , RØD<br />
14 -1<br />
4,6 ⋅10<br />
s<br />
−3<br />
1,00 ⋅10<br />
mm<br />
c) d =<br />
= 2,56·10 -6 n ⋅ λ<br />
, sinφ n =<br />
390<br />
d<br />
−7<br />
1⋅<br />
6,5 ⋅10<br />
m<br />
=> sinφ 1 =<br />
= 0,254 => φ<br />
−6<br />
1 = 15°<br />
2,56 ⋅10<br />
m<br />
Opgave 3<br />
a) U 1 = R 1 · I 1 = 18 Ω · 0,14 A = 2,52 V<br />
b) U 3 = R 3 · I 3 = 33 Ω · 0,071 A = 2,343 V , U 4 = U 0 - U 1 - U 3 = 10,137 V ,<br />
10,137 V<br />
I 4 = I 3 = 0,071 A => R 4 = U 4 /I 4 = = 143 Ω<br />
0,071A<br />
c) U 2 = U 0 - U 1 = 12 V , I 2 = I 1 - I 3 = 0,069 A => P = U 2 · I 2 = 0,86 W<br />
Opgave 4<br />
a) p = F/A => F atmos mod bund = p 0 · A = 173 kN<br />
b) F tyn = mg = 375 kg · 9,82 N/kg = 3,68 kN<br />
F atmos mod top = F atmos mod bund - F tyn - F friktion = 169 kN =><br />
F 169kN<br />
p = = = 99,4 kPa<br />
areal<br />
2<br />
1,70m<br />
c) P atmos mod bund = F · v = 173 kN · 1,20 m/s = 208 kW<br />
d) V = h 0 · A = 8,00 m · 1,70 m 2 = 13,6 m 3<br />
e) ∆V/∆t = v · A = 1,20 m/s · 1,70 m 2 = 2,04 m 3 /s , T = (25,0+273) K = 298 K ,<br />
169kN<br />
p atmos over kabine = F/A = = 99,4 kPa , p · V = n · R · T => p · V/∆t =<br />
2<br />
1,70m<br />
n/∆t · R · T<br />
3<br />
p ⋅ ΔV<br />
/ Δt<br />
99,4 kPa ⋅ 2,04 m / s<br />
=> ∆n/∆t =<br />
=<br />
= 82,0 mol/s<br />
3<br />
R ⋅T<br />
Pa ⋅ m<br />
8,31 ⋅ 298K<br />
mol ⋅ K<br />
<strong>Side</strong> 1 <strong>af</strong> 2
=> ∆m/∆t = ∆n/∆t · M = 122 mol/s · 29,0 g/mol = 2,37 kg/s<br />
Opgave 5<br />
244<br />
a)<br />
94<br />
Pu → 240<br />
92<br />
U + 4 2<br />
He<br />
b) t ½ = ln2/k => k = ln2/t ½ =<br />
ln(2)<br />
7<br />
8,08⋅10<br />
år<br />
= 2,72·10 –16 s –1<br />
A = k · N => N = A/k = 4,56·10 31<br />
A(<br />
t)<br />
ln( )<br />
c) A(t) = A 0 · e –k·t A0<br />
=> t = = 5,37·10 8 år<br />
− k<br />
d) E kin = 7,34·10 –13 J , E kin = ½ ·m ·v 2 , m = 4,00 u = 6,64·10 –27 kg<br />
=> v =<br />
2E kin<br />
m<br />
= 1,49·10 7 m/s<br />
Opgave 6<br />
a) x = 10 cm => F fjeder = k · x = 90 N<br />
b) fjeders vinkel med lodret = φ = tan -1 (b/h) = 53º ,<br />
størrelse <strong>af</strong> fjederkr<strong>af</strong>t i lodret retning = F fjeder,lodret = F fjeder · cos(φ) = 54,0 N<br />
F tyn = m·g = 2,3 kg · 9,82 N/kg = 23 N<br />
F snor = 2·F fjeder,lodret - F tyn = 85,4 N<br />
c) F res = 2·F fjeder - F tyn = 0,12 kN = m·a => a = F res /m = 37,1 m/s 2<br />
d) ∆E pot = m·g·∆h = 6,8 J<br />
E fjeder = ½k·x 2 = 4,5 J<br />
E kin = 2· E fjeder - ∆E pot = 2,2 J = ½mv 2<br />
=> v =<br />
2E kin<br />
m<br />
= 1,39 m/s<br />
<strong>Side</strong> 2 <strong>af</strong> 2