<span>In this problem, we need to solve for Bubba’s mass. To do this, we let A be the area of the raft and set the weight of the displaced fluid with the raft alone as ρwAd1g and ρwAd2g with the person on the raft, </span>where ρw is the density of water, d1 = 7cm, and d2= 8.4 cm. Set the weight of displaced fluid equal to the weight of the floating objects to eliminate A and ρw then solve for m.
<span>ρwAd1g = Mg</span>
ρw<span>Ad2g = (M + m) g</span>
<span>d2∕d1 = (M + m)/g</span>
m = [(d2<span>∕d1)-1] M = [(8.4 cm/7.0 cm) - 1] (600 kg) =120 kg</span>
This means that Bubba’s mass is 120 kg.
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 ![ln[\frac{P_{1} }{P_{2} }]](https://tex.z-dn.net/?f=ln%5B%5Cfrac%7BP_%7B1%7D%20%7D%7BP_%7B2%7D%20%7D%5D)
= <em>mRT</em>₁
[∵<em>pV</em> = <em>mRT</em>]
= (5) (0.287) (272.039) ![ln[\frac{2.0}{1.0}]](https://tex.z-dn.net/?f=ln%5B%5Cfrac%7B2.0%7D%7B1.0%7D%5D)
= 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
Inertia is the property of all matter by which it tends to remain
in constant, uniform motion unless acted on by external force.
Air and water ima air bender