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machinarium

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Please help me quick everyone, thank you in advance.

http://imgkk.com/i/EasJ1d.jpg

a/We have two photogates placed as in the figure. We drop a marble through these at a negligble distance above the upper gate. The upper starts a timer as the ball passes through its beam and the second stop the timer.

Prove that the experimental magnitude of freefall acceleration is given by

[tex]g_{exp}=2\Delta{y}/{(\Delta{t}})^2[/tex]

b/ For the setup and assume that g

c/This time, the upper photogate is placed 0.50cm lower than the first time. But we still drop the marble from the same height as the first time.

What value of g

Based on the problem, I think that [tex]\delta{t}[/tex] is the time when the marble passes through between two gates. But if it were, I couldn't get [tex]g_{exp}=2\Delta{y}/{(\Delta{t}})^2[/tex].

If it were not, I could do like that

We have [tex]{g_{\exp }} = \frac{{\Delta v}}{{\Delta t}}[/tex]

Because v0=0,

Hence,

[tex]g_{exp}=2\Delta{y}/{(\Delta{t}})^2[/tex]

## Homework Statement

http://imgkk.com/i/EasJ1d.jpg

a/We have two photogates placed as in the figure. We drop a marble through these at a negligble distance above the upper gate. The upper starts a timer as the ball passes through its beam and the second stop the timer.

Prove that the experimental magnitude of freefall acceleration is given by

[tex]g_{exp}=2\Delta{y}/{(\Delta{t}})^2[/tex]

b/ For the setup and assume that g

_{exp}=9.81m/s^{2}, what value of [tex]\Delta{t}[/tex] would you expect to measure.c/This time, the upper photogate is placed 0.50cm lower than the first time. But we still drop the marble from the same height as the first time.

What value of g

_{exp}at this time will you determine?## Homework Equations

[tex]a = \frac{{\Delta v}}{{\Delta t}} = \frac{{{v_2} - {v_1}}}{{{t_2} - {t_1}}}[/tex]## The Attempt at a Solution

Based on the problem, I think that [tex]\delta{t}[/tex] is the time when the marble passes through between two gates. But if it were, I couldn't get [tex]g_{exp}=2\Delta{y}/{(\Delta{t}})^2[/tex].

If it were not, I could do like that

We have [tex]{g_{\exp }} = \frac{{\Delta v}}{{\Delta t}}[/tex]

Because v0=0,

Hence,

[tex]g_{exp}=2\Delta{y}/{(\Delta{t}})^2[/tex]

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