Mechanics11/Seite14/Loesung5.4

1.Lösung:

$$\;a = 9.81 \frac{m}{s^{2}} * 10 = 98 \frac{m}{s^{2}}$$

m kann vernachlässigt werden!

$$\;=> F = m * a = 98N$$

$$\;F_Z = m * \omega ^{2} * r$$ $$\;=> \omega ^{2} = \frac{F_Z}{(m * r)} = \frac{98N} {2000m} = 0,049$$ $$\;=> \omega = 0,0221s^{-1}$$ $$\;v = \omega * r = 0,221s^{-1} * 2000m = 442 \frac{m}{s}$$

$$\;v ==> v = 1591 \frac {km} {h}$$

--P 08:14, 8 May 2008 (UTC)

--Kenny 08:14, 8 May 2008 (UTC)

--Chantal 08:14, 8 May 2008 (UTC)

Lösung 2:

$$F_z = a_z = 10 * 9,81 N = 98,1 N$$

r = 2000m

m = 1

$$F_z = m * w^2 * r$$

w = $$\sqrt {\frac {98,1N} {2000m}}$$

w = 0,2215 -1

$$v = w * r = 442,94 m/s = 1594,6 km/h = 1,6 * 10^3 km/h$$

--Gunacke 18:07, 8 May 2008 (UTC)