Oblicz dlugosc wahadla matematycznego,wiedzac ze T=2s,a przyspieszenie ziemskie g=9,81[m/s2].Oblicz l=?
T = 2 s
g = 9,81 m/s²
l = ?
T = 2π√l/g
T² = 4π²l/g
T²*g = 4π²l
l = T²*g/4π²
l = (2s)²* 9,81 m/s²/4 * (3,14)²
l = 4 s² *9,81 m/s² / 4 * 9,86
l ≈ 39,24 m/39,44
l = 0,99 m ≈ 1m
dane:
g = 9,81
szukane:
T = 2TTV(l/g) I^2
4TT^2 *l/g = T^2 I*g
4TT^2 * l = T^2 *g /:4TT^2
l = g *T^2/4TT^2 = 9,81m/s2 *(2s)^2/4*(3,14)^2 = 9,81 *4/4*9,86 m = 9,81/9/86 m
= 0,99 m
l = ok.1m
=========
Odp.Szukana długość wahadła wynosi ok.1m.
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T = 2 s
g = 9,81 m/s²
l = ?
T = 2π√l/g
T² = 4π²l/g
T²*g = 4π²l
l = T²*g/4π²
l = (2s)²* 9,81 m/s²/4 * (3,14)²
l = 4 s² *9,81 m/s² / 4 * 9,86
l ≈ 39,24 m/39,44
l = 0,99 m ≈ 1m
dane:
T = 2 s
g = 9,81
szukane:
l = ?
T = 2TTV(l/g) I^2
4TT^2 *l/g = T^2 I*g
4TT^2 * l = T^2 *g /:4TT^2
l = g *T^2/4TT^2 = 9,81m/s2 *(2s)^2/4*(3,14)^2 = 9,81 *4/4*9,86 m = 9,81/9/86 m
= 0,99 m
l = ok.1m
=========
Odp.Szukana długość wahadła wynosi ok.1m.