Aflevering om rumfang matematik
Afleveringen om rumgeometri af Mike J og Jonas J.
Indholdsfortegnelse |
Opg 242
- 5*10*1,8*1000 = 90000 L.
Opg 243 (skal ikke laves)
- 3³ = 27 cm³.
- 3²*6 = 54 cm².
Opg 244
- (15)^(1/3) = 2,47 cm.
Opg 246
- Gd =(3,5+1,5)/2-0,3 = 2,2 m (Gennemsnits dybde)
- V = Gd * 25² = 1375 m³ (Mængde af vand)
- P = 300+350 = 650 L/min (Pumpe kraft)
- T = (V*1000)/P = 2115 min
- 35 timer og et kvarter.
Opg 247
- (2²*3)+(2²*3/2) = 18 m³.
Opg 250
- 2²*7*7500/100³ = 0,21 kg.
- 0,21-((1,5*3*2/2+3,5)*7500/100³) = 0,15 kg.
Opg 251
- 6,5*100*(20/2)²*PI*7800/100³ = 1592 kg.
- 20*PI*6,5*100 = 40840,7 cm².
Opg 255
- (((3/2)²*PI)-(((3-0,2*2)/2)²*PI))*2,8*100*8600/100³ = 4,24 kg.
Opg 257
- 1/3*8²*10 = 213,3 cm³
- ((8+11.5+11.5)(8+11.5-11.5)(11.5+11.5-8)(11.5+11.5-8))^(1 / 2) = 236.220236
Opg 258
- v = 1/3*(3*5²*Sin(60))*(8²-5²)^(1/2) = 135.204cm³.
- o = (((5+5+5)(5+5-5)(5+5-5)(5+5-5))^(1/2))/4*6+(((5+8+8)(5+8-8)(8+8-5)(8+5-8))^(1/2))/4*6 = 178,94 cm².
Opg 261
- v = (14²-(8/2)²)^(1/2)*1/3*(8/2)²*PI = 224,68 cm³.
- O = PI*(8/2)*14 = 175,84 cm².
- (360*(8/2))/14 = 102,8 grader.
Opg 264
- v = PI/3*250*((240/2)²+(180/2)²+(240/2)*(180/2)) +PI/3*(80+40-60)*((240/2)²+((60-40)/80*(240/2-40/2))²+(240/2)*((60-40)/80*(240/2-40/2))) = 9845470mm³.
- L = 9845470/100³ = 9,845470 L
Opg 266
- AB² = 2*PI*1*1*3 = 18,8
- AC² = 2*PI*1*((4-1)²+(3-1)²)^(1/2)*((4-1)/2+1) = 56,6
- BC² = 2*PI*1*((4-1)²+(3-4)²)^(1/2)*((4-1)/2+1) = 49,7
- O = 56,6+49,7+18,8 = 125,1
Opg 275
- V = PI/6*1,8²*(3*16-2*1,8) = 75,3 cm³.
- O = (PI*1,8*16)+((PI*1,8*16)/PI-1,8²)*PI = 170,8 cm².
Opg 277
- V = 4/3*PI*(100/2)³ = 523598 mm³
- h=100-(100^2-50^2)^(1/2) = 13,3974596 mm
- Vv = PI/6*h*(3*(50/2)²+h²)*2+PI*(50/2)²*(100^2-50^2)^(1/2) = 198867,774 mm³
- V% = Vv/V*100 = 37,98 %