Two boats - Boat A and Boat B - are anchored a distance of 24 meters apart. The incoming water waves force the boats to oscillat
1 answer:
Answer:
wavelength = 24 m
Period = 10 s
f = 0.1 Hz
Amplitude = 4 m
Explanation:
Wavelength:
Since the boats are at crest and trough, respectively at the same time. Hence, the horizontal distance between them is the wavelength of the wave:
<u>wavelength = 24 m</u>
Period:
The period is given as:
<u>Period = 10 s</u>
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Frequency:
The frequency is given as:
<u>f = 0.1 Hz</u>
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Amplitude:
Amplitude will be half the distance between extreme points, that is, crest and trough:
Amplitude = 8 m/2
<u>Amplitude = 4 m</u>
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Explanation:
We need to stablish that there is 3 states at this problem. At the firts one, water is compressed liquid and the conditions for this state are: P1=100KPa,T1=20°C,V1=0.5m^3. From the compressed liquid chart and using extrapolation, we can get: v1=vf1=0.0010018 (m^3/Kg) and u1=uf1=83.95(KJ/kg). Now we can find the mass of water at the state 1 as: Then the liquid water is heated at a rate of 0.85KW, and its volume increase, while work is done by the system at the boundary, we can assume that the pressure remains constant throughout the entire process. At the second state the water is saturated liquid and the conditions are: P2=100KPa, T2=Tsat=99.63°C, v2=vf2=0.001043(m^3/Kg) and u2=uf2=417.36(KJ/Kg). Now we can find the work as:
. (a) After that we need to do an energy balance for process 1-2 and get: U=Q-W or
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.(b) Then continue heat transfer to the cooking pot and results in phase change getting vapor at 99.63°C. At the final state or third state the mass is zero because all liquid was evaporated and the initial mass at this state is the same for the second state: 0.5 (Kg) and doing an energy balances results in:
, but m3=0, now solving for t we can get the time required for all of the water to evaporate as:
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. Now replacing every value in the time equation we get:
