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kkurt [141]
3 years ago
11

Given an array of words representing your dictionary, you test words to see if it can be made into another word in dictionary. T

his will be done by removing characters one at a time. Each word represents its own first element of its string chain, so start with a string chain length of 1. Each time you remove a character, increment your string chain by 1. In order to remove a character the resulting word must be in your original dictionary. Your goal is to determine the longest string chain available for a given dictionary. for example, given a dictionary [a, and, ab, bear] the word and could be reduced to an and the word an to a. The single character a can not be reduced to any further as the null string is not in the dictionary. This would be the longest string chain, having a length 3. The word bear can not be reduced at all.

Engineering
1 answer:
katen-ka-za [31]3 years ago
8 0

Answer:

Detailed solution is given below:

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4 years ago
An automotive fuel cell consumes fuel at a rate of 28m3/h and delivers 80kW of power to the wheels. If the hydrogen fuel has a h
EastWind [94]

Answer:

The efficiency of this fuel cell is 80.69 percent.

Explanation:

From Physics we define the efficiency of the automotive fuel cell (\eta), dimensionless, as:

\eta = \frac{\dot W_{out}}{\dot W_{in}} (Eq. 1)

Where:

\dot W_{in} - Maximum power possible from hydrogen flow, measured in kilowatts.

\dot W_{out} - Output power of the automotive fuel cell, measured in kilowatts.

The maximum power possible from hydrogen flow is:

\dot W_{in} = \dot V\cdot \rho \cdot L_{c} (Eq. 2)

Where:

\dot V - Volume flow rate, measured in cubic meters per second.

\rho - Density of hydrogen, measured in kilograms per cubic meter.

L_{c} - Heating value of hydrogen, measured in kilojoules per kilogram.

If we know that \dot V = \frac{28}{3600}\,\frac{m^{3}}{s}, \rho = 0.0899\,\frac{kg}{m^{3}}, L_{c} = 141790\,\frac{kJ}{kg} and \dot W_{out} = 80\,kW, then the efficiency of this fuel cell is:

(Eq. 1)

\dot W_{in} = \left(\frac{28}{3600}\,\frac{m^{3}}{s}\right)\cdot \left(0.0899\,\frac{kg}{m^{3}} \right)\cdot \left(141790\,\frac{kJ}{kg} \right)

\dot W_{in} = 99.143\,kW

(Eq. 2)

\eta = \frac{80\,kW}{99.143\,kW}

\eta = 0.807

The efficiency of this fuel cell is 80.69 percent.

3 0
3 years ago
Sophia is designing a new welding shop for the local high school. Where should the compressed gas and fuel cylinders be stored?
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4 years ago
An archer releases an arrow toward a target. The arrow travels 166 meters in 2 seconds. The speed of the arrow is
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Explanation:

speed= distance/time

=166/2

=83m/s

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