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statuscvo [17]
3 years ago
13

Help thank you <3 :DDD

Engineering
1 answer:
Vika [28.1K]3 years ago
4 0

Answer:

Explanation:

its B

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WARNING:<br><br> when people put links in the answer it is a virus DO NOT DOWNLOAD IT
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There are 20 forging presses in the forge shop of a small company. The shop produces batches of forgings requiring a setup time
Aleksandr-060686 [28]

Answer:

Considering the guidelines of this exercise.

The pieces produced per month are 504 000

The productivity ratio is 75%

Explanation:

To understand this answer we need to analyze the problem. First of all, we can only produce 2 batches of production by the press because we require 3 hours to set it up. So if we rest those 6 hours from the 8 of the shift we get 6, leaving 2 for an incomplete bath. So multiplying 2 batches per day of production by press we obtain 40 batches per day. So, considering we work in this factory for 21 days per month well that makes 40 x 21  making 840 then we multiply the batches for the pieces 840 x 600 obtaining 504000 pieces produced per month. To obtain the productivity ratio we need to divide the standard labor hours meaning 6 by the amount of time worked meaning 8. Obtaining 75% efficiency.

4 0
3 years ago
1. Band saw lower wheel does not require a guard * true or false 2. Band saw upper guide should be adjusted to within 1/8" of th
Blizzard [7]

Answer:

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4 0
2 years ago
A block of ice weighing 20 lb is taken from the freezer where it was stored at -15"F. How many Btu of heat will be required to c
Rus_ich [418]

Answer:

Heat required =7126.58 Btu.

Explanation:

Given that

Mass m=20 lb

We know that

1 lb =0.45 kg

So 20 lb=9 kg

m=9 kg

Ice at -15° F and we have to covert it at 200° F.

First ice will take sensible heat at up to 32 F then it will take latent heat at constant temperature and temperature will remain 32 F.After that it will convert in water and water will take sensible heat and reach at 200 F.

We know that

Specific heat for ice C_p=2.03\ KJ/kg.K

Latent heat for ice H=336 KJ/kg

Specific heat for ice C_p=4.187\ KJ/kg.K

We know that sensible heat given as

Q=mC_p\Delta T

Heat for -15F to 32 F:

Q=mC_p\Delta T

Q=9\times 2.03(32+15) KJ

Q=858.69 KJ

Heat for 32 Fto 200 F:

Q=mC_p\Delta T

Q=9\times 4.187(200-32) KJ

Q=6330.74 KJ

Total heat=858.69 + 336 +6330.74 KJ

Total heat=7525.43 KJ

We know that 1 KJ=0.947 Btu

So   7525.43 KJ=7126.58 Btu

So heat required to covert ice into water is 7126.58 Btu.

8 0
3 years ago
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