Attached image shows how to get
$$dE = -Tdx + E\frac{dx} {x}$$
This is interesting. But I'm not sure how this approach helps in solving the problem ;-)
To solve this problem, these are the key equations to be resolved.
$$\frac {dm} {dx} = \frac {M} {L} $$.... get m(x)
$$T=ma ⇒ \frac {dT} {dx} = 0 = m \frac {da} {dx} + a \frac {dm} {dx} $$.... get a(x)
$$a = \frac {dv} {dt} = \frac {dv} {dx} \frac {dx} {dt} = v \frac {dv} {dx} ⇒ a dx = v dv...
Hi, everything you did was correct, except that m_A should be not included in this free body diagram analysis. Review this. That's causing error in your calculation.
The confusion is in the question itself.
If gravity is 9.2 m/s^2 (as stated in the question), then satellite distance from earth center is 6580km (verify yourself!). And that distance should not be called altitude :-(
If altitude is really 6580km, g is about 2.38 m/s^2.
In your experiment, s = (1/2)g t2
After your conversion and plot, your graph is similar to following equation.
y = (1/2)g x .... general equation of a straight line.
Slope is therefore (1/2)g, .... there's nothing wrong, right?
1) Correct
2) Wrong. Check yr answer again!
3) What are the generic rules for below:
sin(π + θ) = ?
cos(π + θ) = ?
Apply above result to below to see relation with 1.
Cy = C sin(ϒ) = C sin(π + δ) = ?
Cx = C cos(ϒ) = C cos(π + δ) = ?
4) What are the generic rules for
sin(2π + θ) = ?
cos(2π +...
Exactly. I feel the question is incomplete. When I work on this question, I assume piston is not moved. If so, how would you expect the result? Kindly share.
There is no more conditions given to this question.
I believed that additional mercury will compress the gas inside cylinder, therefore gas pressure increases. As a result, both mercury levels will increase. And the difference of the mercury levels is higher than 0.2m.