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Alex17521 [72]
3 years ago
7

3. (20 points) Suppose we wish to search a linked list of length n, where each element contains a key k along with a hash value

h(k). Each key is a long character string. How might we take advantage of the hash values when searching the list for an element with a given key?
Engineering
1 answer:
Gre4nikov [31]3 years ago
6 0

Answer:

Alternatively we produce a complex (hash) value for key which mean that "to obtain a numerical value for every single string" that we are looking for.  Then compare that values along the range of list, that turns out be numerical values so that comparison becomes faster.

Explanation:

Every individual key is a big character  so to compare every keys, it is required to conduct a quite time consuming string reference procedure at every node. Alternatively we produce a complex (hash) value for key which mean that "to obtain a numerical value for every single string" that we are looking for.  Then compare that values along the range of list, that turns out be numerical values so that comparison becomes faster.

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A piston–cylinder assembly contains air, initially at 2 bar, 300 K, and a volume of 2 m3. The air undergoes a process to a state
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Work, W = 277.269kJ

Internal energy, Q = 277.269kJ

<u>Explanation:</u>

Given-

Pressure, P1 = 2 bar

Temperature, T1 = 300K

Volume, V1 = 2m³

P2 = 1 bar

PV = constant

Let,

mass in kg be m

Work in kJ be W

Heat transfer in kJ be Q

R' = 8.314 kJ/kmolK

Mass of air, Mair = 28.97 kg/kmol

R = 0.289 kJ/kgK

We know,

PV = mRT

m = \frac{P_1V_1}{RT_1}

m = 5.65kg

To calculate V₂:

PV = constant = P₁V₁ = P₂V₂

P₁V₁ = P₂V₂

V_2 = \frac{P_1V_1}{P_2}

V₂ = 4m³

To calculate the work:

P₁V₁ = C

P₁ = C/ V₁

W = \int\limits^V_V {pdV} \,

where limit is V₁ to V₂

W =  \int\limits^V_V {\frac{c}{v} } \, dV \\\\W = C\int\limits^V_V {v^-^1} \, dV\\ \\W = P_1V_1 (ln\frac{V_2}{V_1} ) \\\\W = (2 bar) (2m^3) (ln\frac{4m^3}{2m^3}) (\frac{10^5 N/m^2}{1 bar}) \\\\W = 277.259kJ

To calculate heat transfer:

Q - W = Δu

Q - W = m (u₂ - u₁)

Q = W + m (u₂ - u₁)

Q = W + m X cv X (T₂ - T₁)

Since, T₁ ≈ T₂

There is no change of internal energy.

W = Q

Q = 277.269kJ

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