Suppose that a wireless link layer using a CSMA-like protocol backs off 1ms on average. A packet’s link and physical layer heade
1 answer:
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Answer:
a.
b.
Explanation:
Hello,
a. In this case, the following energy balance must be stated:
Δ
,
Now, at 30°C which is both the initial and inlet temperature of nitrogen, the referred enthalpy has a value of 8,815 J/mol (Cengel's book), thus, one can say that:
In such a way, we could factor as follows:
Now, since is defined as the addition between the initial mass and the inlet mass of nitrogen, one sum up that:
Thus, since the initial enthalpy, computed at 30°C remains constant, it means that the final temperature remains constant, so:
This is evident due to the fact that the inlet nitrogen has the same initial nitrogen and no heat is transferred.
b. Considering the final temperature, one proceeds to compute the final mass as follows:
Now, as long as the temperature changes to 20°C, the new pressure is computed as:
Answer:
a) 1.16 m/s
b) 1/216000
c) (√15)/6480000
Explanation:
The parameters given are;
Length of boat prototype, lp = 30 m
Speed of boat prototype = 9 m/s
Length of boat model, lm= 0.5 m
a) lm/lp = 0.5/30 = 1/60 = ∝
(vm/vp) = ∝^(1/2) = √∝ = (1/60)^(1/2)
vm = 9 × (1/60)^(1/2) = 1.16 m/s
b) The ratio of the model to prototype drag, Fm/Fp, is given as follows;
Fm/Fp = (vm/vp)²×(lm/lp)² = ∝³
Fm/Fp = (1/60)³ = 1/216000
c) The ratio of the model to prototype power pm/p = (Fm/Fp) × (vm/vp) = ∝³×√∝
The ratio of the model to prototype power pm/p = √(1/60) × (1/60)³
pm/p = √(1/60) × (1/60)³ = (√15)/6480000