Respuestas
Respuesta dada por:
1
1.I. Fe=kx Fe=Fg
⇒ kx = mg
k = mg/x →![k= \frac{0.25*9.8}{0.056} = 43.75 kg.s^-^2 k= \frac{0.25*9.8}{0.056} = 43.75 kg.s^-^2](https://tex.z-dn.net/?f=k%3D++%5Cfrac%7B0.25%2A9.8%7D%7B0.056%7D++%3D+43.75+kg.s%5E-%5E2)
II. f=n(oscilaciones)/tiempo
f₁=8/9.52= 0.840 s-1
f₂=8/9.72= 0.823 s-1
f₃=8/9.47= 0.845 s-1
III. Fe=Fg
43.75(0.092) = 9.8m
m = 0.41kg= 410.7g
2.
T:periodo
T = 2π√(0.4/9.8) = 1.27 s
I. f = 1/T
f = 1/1.27 = 0.787 s-1
II. 2 oscilaciones = 2T =2.54 s
3.![PV=RTn PV=RTn](https://tex.z-dn.net/?f=PV%3DRTn)
126200 Pa = 1.25 atm 138400Pa = 1.37 atm
15cm3 = 15 mL 28°C = 301°K
I. 1.25*15*10^-3 = 0.082*301*n
n = 7.6*10^-4 mol = 0.76 mmol
II. 1.37*V = 0.082*301*7.6*10^-4
V = 0.0137 L = 13.7 mL
⇒ kx = mg
k = mg/x →
II. f=n(oscilaciones)/tiempo
f₁=8/9.52= 0.840 s-1
f₂=8/9.72= 0.823 s-1
f₃=8/9.47= 0.845 s-1
III. Fe=Fg
43.75(0.092) = 9.8m
m = 0.41kg= 410.7g
2.
T = 2π√(0.4/9.8) = 1.27 s
I. f = 1/T
f = 1/1.27 = 0.787 s-1
II. 2 oscilaciones = 2T =2.54 s
3.
126200 Pa = 1.25 atm 138400Pa = 1.37 atm
15cm3 = 15 mL 28°C = 301°K
I. 1.25*15*10^-3 = 0.082*301*n
n = 7.6*10^-4 mol = 0.76 mmol
II. 1.37*V = 0.082*301*7.6*10^-4
V = 0.0137 L = 13.7 mL
ronaldofcr:
Muchas gracias, :)
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