Question

Oblicz okres i częstotliwość obrotu koła roweru jadacego z predkościa 18km/h i śr.13 cal

Answer 1

1 cal = 2,54cm

 

13cali ≈ 33cm

 

d = 33cm

r = d/2

r = 33cm/2 = 16,5cm

v = 18km/h = 5m/s

 

s = Obw = 2πr

s = 2 x 3,14 x 16,5cm ≈ 104cm

t = T = ?

s = vt => t = s/v

T = 1,04m/5m/s = 0,21s

 

f = 1/T

f = 1/0,21s = 4,76Hz

 

 

 

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Litterarum radices amarae sunt, fructus iucundiores

 

Pozdrawiam :)

Answer 2

Ze wzorów:

 

T = t/n

f = 1/T

 

Koło =   2,54 cm*13cal = 33,02cm = 3,3m 

 

zamieniamy 18km/h na m/s 

 

1000*18km/3600 = 18000/3600 = 5m/s

 

 

V = 2πr /T

V * T = 2πr

T = 2πr/V

T = 2 *3,14 *16,5 cm /5m/s

T ≈ 104cm ≈ 0,21s

f = 1/T

f = 1/T

f = 1/0,21s

= 4,76Hz