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jonny [76]
3 years ago
10

12. Dies are turned using a special tool called a/an

Engineering
1 answer:
PIT_PIT [208]3 years ago
3 0

Answer:

the answer is die stock

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To test the effects of a new fertilizer, 100 plots were divided in half. Fertilizer A is randomly applied to one half, and B to
Kruka [31]

Answer:

(b) a matched pairs experiment.

Explanation:

Matched pairs experiment are a special case of a random block design and is usually used when the experiment only has two conditions. In this case the conditions are the different fertilizers.

7 0
3 years ago
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Which of the following is typical of the selection of survey participants? They can indicate partiality toward a group. They mus
givi [52]

Answer:

Option D

Explanation:

In general a sample collected from a set of naturally distributed data set must also follow the bell curve or natural distribution

In other way round it can be said that sample set must have the same characteristics as that of the main set and hence a sample must be representative of the population.

Option D is correct

3 0
3 years ago
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What information on a piping arrangement drawing gives the vertical placement of an item?
Yakvenalex [24]

Answer:

c) Elevations

Explanation:

Elevations in pipe systems design provide the views necessary in presenting a pipe isometrics indicating the pipe length, wideness of a line and the height of a pipe.The drafts person obtains the true north coordinates in the desired points where the pipeline is to pass and determines the required vertical grade where all vertical measurements will be referred.

4 0
4 years ago
Helium is used as the working fluid in a Brayton cycle with regeneration. The pressure ratio of the cycle is 8, the compressor i
Temka [501]

Answer:

Explanation:

Find the temperature at exit of compressor

T_2=300 \times 8^{\frac{1.667-1}{1.667} }\\=689.3k

Find the work done by the compressor

\frac{W}{m} =c_p(T_2-T_1)\\\\=5.19(689.3-300)\\=2020.4kJ/kg

Find the actual workdone by the compressor

\frac{W}{m} =n_c(\frac{W}{m} )\\\\=1 \times 2020.4kJ/kg

Find the temperature at exit of the turbine

T_4=\frac{1800}{8^{\frac{1.667-1}{1.667} }} \\\\=787.3k

Find the actual workdone by the turbine

1 \times 5.19 (1800-783.3)\\=5276.6kJ/kg

Find the temperature of the regeneration

\epsilon = \frac{T_5-T_2}{T_4-T_2} \\\\0.75=\frac{T_5-689.3}{783.3-689.3} \\\\T_5=759.8k

Find the heat supplied

Q_i_n=c_p(T_3-T_5)\\\\=5.19(1800-759.8)\\\\=5388.2kJ/kg

Find the thermal efficiency

n_t_h=\frac{W_t-W_c}{Q_i_n} \\\\=\frac{5276.6-2020.4}{5388.2} \\\\n_t_h=60.4

60.4%

Find the mass flow rate

m=\frac{W_net}{P} \\\\\frac{60 \times 10^3}{5276.6-2020.4} \\\\=18.42

Find the actual workdone by the compressor

\frac{W_c}{m} =\frac{(\frac{W}{m} )}{n_c} \\\\=\frac{2020.4}{0.8} \\\\=2525.5kg

Find the actual workdone by the turbine

\frac{W_t}{m} =n_t(\frac{W}{m} )\\\\=0.8 \times5.19(1800-783.3)\\\\=4221.2kJ/kg

Find the temperature of the compressor exit

\frac{W_t}{m} =c_p(T_2_a-T_1)\\2525.5=5.18(T_2_a-300)\\T_2_a=787.5k

Find the temperature at the turbine exit

4221.2=5.18(1800-T_4_a)\\\\T_4_a=985k

Find the temperature of regeneration

\epsilon =\frac{T_5-T_2}{T_4-T_2}\\\\0.75=\frac{T_5-787.5}{985-787.5}\\\\T_5=935.5k

6 0
3 years ago
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Which activity promotes the preservation of species
miv72 [106K]

Answer:

A

Explanation:

8 0
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