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merkurus
8 months ago

Aufgabe 2
Kugel Volumen Vk
Vk = (4/3) * pi * rk³
Vk = (4/3) * pi() * 4^3
Vk = 268,082573106329 mm³

Zylinder Höhe bzw. Dicke hz
Vz = rz² * pi * hz
hz = V / ( rz² * pi() )
hz = 268,082573106329 / ( 500^2 * pi() )
hz = 0,000341333333 mm
Die Dicke des Ölflecks ist 0,000341333 mm

Aufgabe 3a
V = G * h
V = 3350 * 90
V = 301500 m³
Fassungsvermögen ist 301500 m³

Aufgabe 3b
V = (4/3) * pi * r³
r³ = V / ( (4/3) * pi() )
r = Dritte Wurzel( V / ((4/3) * pi) )
r = Dritte Wurzel( 301500 / ((4/3) * pi()) )
r = 41,597405 m

d = r * 2
d = 41,597405 * 2
d = 83,19481 m
Kugelgasbehälter d = 83,195 m

Halbrecht
8 months ago

At the 1st task I have the radius and the approach VolumeKegel = VolumeKegelsmall + VolumeKugel

(1)

It fits into the cone with radius 4 and height 8 a volume of

V = 1/3 *

.

The remaining 10%

These are 0.1* V = V_Rest

.

Now

V_Rest = 4/3 *

relocation to r

r = 3*V_Rest/4pi