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3 years ago

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Suppose you were doing an A* search and there were a number of different ways that you could compute an admissible heuristic val

ue. Suppose that some were very cheap to compute but also very inaccurate estimators while others were very expensive to compute but were very accurate estimators. How would you determine which computation to use for a given application?

1 answer:

Elodia [21]3 years ago

5 0

Answer:

Explanation:

First of all,we need to know what kind of research and how important it is before determining which computation method to use .

Then compare end result of research and cost of expensive methods .

Then we should opt for the best expensive method to be used for the given application.

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Explain the four criteria for proving the correctness of a logical pretest loop construct of the form "while B do S end". And pr

evablogger [386]

Answer:

Check the explanation

Explanation:

The loop invariant has to satisfy some amount of requirements to be of good use. Another complex factor as to why a loop is the question of loop termination. A loop that doesn’t terminate can’t invariably be correct, and in fact the computation in whatever form amounts to nothing. The total axiomatic description of a while construct will have to involve all of the following to be true, in which I is the loop invariant:

P => I

{I and B} S {I}

(I and (not B)) => Q

Then the loop terminates

6 0

2 years ago

Two sites are being considered for wind power generation. On the first site, the wind blows steadily at 7 m/s for 3000 hours per

kirill [66]

Solution :

Given :

$V_1 = 7 \ m/s$

Operation time, $T_1$ = 3000 hours per year

$V_2 = 10 \ m/s$

Operation time, $T_2$ = 2000 hours per year

The density, ρ = $1.25 \ kg/m^3$

The wind blows steadily. So, the K.E. = $(0.5 \dot{m} V^2)$

$= \dot{m} \times 0.5 V^2$

The power generation is the time rate of the kinetic energy which can be calculated as follows:

Power = $\Delta \ \dot{K.E.} = \dot{m} \frac{V^2}{2}$

Regarding that $\dot m \propto V$ . Then,

Power $\propto V^3$ → Power = constant x $V^3$

Since, $\rho_a$ is constant for both the sites and the area is the same as same winf turbine is used.

For the first site,

Power, $P_1= \text{const.} \times V_1^3$

$P_1 = \text{const.} \times 343 \ W$

For the second site,

Power, $P_2 = \text{const.} \times V_2^3 \ W$

$P_2 = \text{const.} \times 1000 \ W$

5 0

2 years ago

A direct contact heat exchanger (where the fluid mixes completely) has three inlets and one outlet. The mass flow rates of the i

lara31 [8.8K]

Answer:

Enthalpy at outlet=284.44 KJ

Explanation:

$m_1=1 Kg/s,m_2=1.5 Kg/s,m_3=22 Kg/s$

$h_1=100 KJ/Kg,h_2=120 KJ/Kg,h_3=500 KJ/Kg$

We need to Find enthalpy of outlet.

Lets take the outlet mass m and outlet enthalpy h.

So from mass conservation

$m_1+m_2+m_3=m$

m=1+1.5+2 Kg/s

m=4.5 Kg/s

Now from energy conservation

$m_1h_1+m_2h_2+m_3h_3=mh$

By putting the values

$1\times 100+1.5\times 120+2\times 500=4.5\times h$

So h=284.44 KJ

4 0

3 years ago

Liquefied Natural Gas (LNG) is a natural gas in its liquid form that is clear, colorless, odorless, non-corrosive, and non-toxic

ZanzabumX [31]

Liquefied Natural Gas (LNG) can be defined as a natural gas which in liquid form appear clear and colorless. It is odorless, non-toxic, and non-corrosive. Therefore, the given statement is A) True.

LNG or Liquified Natural Gas is a fossil fuel that is produced after the compression of organic matter in the form of algae and phytoplankton.
LNG consists of 95% methane gas.
The combustion of LNG produces carbon dioxide and water vapors.
It burns with a least pollution thus called as cleanest fossil fuel.
The liquefaction of the natural gas takes place at -160 degree Celsius. The liquefaction of the gas causes it to transport easily in gas tanks.
LNG is colorless, and clear.
LNG does not possess any smell and it is non-corrosive to metallic tanks.
LNG is also non-toxic.

Learn more about natural gas:

brainly.com/question/12200462

6 0

2 years ago

Determine the hydraulic radius for the following rectangular open channel width =23m water depth =3m

Romashka-Z-Leto [24]

Answer:

2.379m

Explanation:

The width = 23m

The depth = 3m

The radius is denoted as R

The wetted area is = A

The perimeter perimeter = P

Hydraulic radius

R = A/P

The area of a rectangular channel

= Width multiplied by Depth

A = 23x3

A = 69m²

Perimeter = (2x3)+23

P = 6+23

P= 29

Hydraulic radius R = 69/29

= 2.379m

This answers the question

Thank you!

8 0

2 years ago