A uniform ladder of length L and weight w is leaning against a vertical wall. The coefficient of static friction between the lad
1 answer:
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Answer:
The work and heat transfer for this process is = 270.588 kJ
Explanation:
Take properties of air from an ideal gas table. R = 0.287 kJ/kg-k
The Pressure-Volume relation is <em>PV</em> = <em>C</em>
<em>T = C </em> for isothermal process
Calculating for the work done in isothermal process
<em>W</em> = <em>P</em>₁<em>V</em>₁
= <em>mRT</em>₁ [∵<em>pV</em> = <em>mRT</em>]
= (5) (0.287) (272.039)
= 270.588 kJ
Since the process is isothermal, Internal energy change is zero
Δ<em>U</em> =
From 1st law of thermodynamics
Q = Δ<em>U </em>+ <em>W</em>
= 0 + 270.588
= 270.588 kJ
Answer:
Explanation:
Given that,
h(t) = -9.8t² / 2 + 125t + 500
for t > 0.
At t = 0, the rocket is at height h = 500m, at a velocity of Vo = 125m/s.
We want to find the maximum height reached by rocket
Using mathematics maxima and minima
let find the turning point when dh/dt = 0
dh/dt = -9.8t + 125
dh / dt = 0 = -9.8t + 125
9.8t = 125
t = 125 / 9.8
t = 12.76s
Let find the turning point to know if this time t = 12.76 is maximum or minimum point
Let find d²h / dt²
d²h / dt² = -9.8
Since, d²h/dt² < 0, then, at t = 12.76s is the maximum points.
Then, the maximum height reached is
h = -9.8t² / 2 + 125t + 500
h = -9.8(12.76)² / 2 + 125(12.76) + 500
h = -797.80 + 1595 + 500
h = 1297.2 m
The maximum height reached is 1297.2 m
From the attachment, the maximum height is 1297.2m at t = 12.76sec.
Comment, the result are the same for both the analysis aspect and the graphical aspect.
