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Setler79 [48]
3 years ago
5

Given the strings s1 and s2, not necessarily of the same length, create a new string consisting of alternating characters of s1

and s2 (that is, the first character of s1 followed by the first character of s2, followed by the second character of s1, followed by the second character of s2, and so on. Once the end of either string is reached, no additional characters are added. For example, if s1 contained "abc" and s2 contained "uvwxyz", then the new string should contain "aubvcw". Assign the new string to the variable s3.
Engineering
1 answer:
andrew11 [14]3 years ago
4 0

Answer:

THE ANSA IZ A

Explanation:

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An FPC 4 m2 in area is tested during the night to measure the overall heat loss coefficient. Water at 60 C circulates through th
sp2606 [1]

Answer:

<em> - 14.943 W/m^2K  ( negative sign indicates cooling ) </em>

Explanation:

Given data:

Area of FPC = 4 m^2

temp of water = 60°C

flow rate = 0.06 l/s

ambient temperature = 8°C

exit temperature = 49°C

<u>Calculate the overall heat loss coefficient </u>

Note : heat lost by water = heat loss through convection

m*Cp*dT  = h*A * ( T - To )

∴ dT / T - To = h*A / m*Cp  ( integrate the relation )

In ( \frac{49-8}{60-8} ) =  h* 4 / ( 0.06 * 10^-3 * 1000 * 4180 )

In ( 41 / 52 ) = 0.0159*h

hence h = - 0.2376 / 0.0159

              = - 14.943  W/m^2K  ( heat loss coefficient )

7 0
3 years ago
The structure of a house is such that it loses heat at a rate of 4500kJ/h per °C difference between the indoors and outdoors. A
adelina 88 [10]

Answer:

15.24°C

Explanation:

The quality of any heat pump pumping heat from cold to hot place is determined by its coefficient of performance (COP) defined as

COP=\frac{Q_{in}}{W}

Where Q_{in} is heat delivered into the hot place, in this case, the house, and W is the work used to pump heat

You can think of this quantity as similar to heat engine's efficiency

In our case, the COP of our heater is

COP_{heater} = \frac{\frac{4500\ kJ}{3600\ s} *(T_{house}-T_{out})}{4\ kW}

Where T_{house} = 24°C and T_{out} is temperature outside

To achieve maximum heating, we will have to use the most efficient heat pump, and, according to the second law of thermodynamics, nothing is more efficient that Carnot Heat Pump

Which has COP of:

COP_{carnot}=\frac{T_{house}}{T_{house}-T_{out}}

So we equate the COP of our heater with COP of Carnot heater

\frac{1.25 *(T_{house}-T_{out})}{4}=\frac{T_{house}}{T_{house}-T_{out}}

Rearrange the equation

\frac{1.25}{4}(24-T_{out})^2-24=0

Solve this simple quadratic equation, and you should get that the lowest outdoor temperature that could still allow heat to be pumped into your house would be

15.24°C

4 0
3 years ago
wo companies, Ajax Co. and Boho Inc., were negotiating a merger. In the course of the negotiations, an Ajax representative told
boyakko [2]

Answer:

Yes

Explanation:

If the Ajax representative fails to correct the previous statement this can cause misrepresentation.

4 0
3 years ago
Comparación de hipotecas Los Chos
aleksklad [387]

Answer:

I don't understand the language French sorry can't answer

3 0
3 years ago
About ceramics: Only can be optically opaque or semi-transparent. a) True b)-False
julia-pushkina [17]

Answer: True

Explanation: Ceramics have the property that when the band gap present between the atoms are larger than the light energy then the tend to become opaque because the light scattering is caused . They also show the property of being translucent when there are chances of the light to get a path through the surface of ceramic so they get the light at some parts e.g.porcelain .Therefore the statement given is true that ceramics can be optically opaque or semi-transparent(translucent).

6 0
3 years ago
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