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forsale [732]
3 years ago
11

Ensure at least ___ distance around fire sprinkler heads, safety showers, eyewash units, and heating and cooling units to ensure

proper operation.
Engineering
1 answer:
vampirchik [111]3 years ago
5 0

90 inches

Explanation:

According to OSHA requirement, the distance around safety showers and eyewash should be between 82-96 inches off the flow. This will allow for maximum diameter of spray.

Learn More

Safety distance around safety showers:brainly.com/question/11123362

Keywords: distance, fire sprinkler head, safety showers, eyewash units,heating and cooling units

#LearnwithBrainly

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Yeah that is important
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Write a function "funthree" that will print a box of characters. The function will always receive as the first input argument th
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4 years ago
Nancy ate a 500 Cal lunch. Neglecting efficiency issues (i.e., assuming 100% conversion of energy to work), to what height could
VladimirAG [237]

Answer:

4265.04\ \text{m}

2.38\times 10^{10}\ \text{W}

Explanation:

PE = Energy of food = 500 cal = 500\times4184=2.092\times10^6\ \text{J}

m = Mass of object = 50 kg

g = Acceleration due to gravity = 9.81\ \text{m/s}^2

Potential energy of food is given by

PE=mgh\\\Rightarrow h=\dfrac{PE}{mg}\\\Rightarrow h=\dfrac{2.092\times 10^6}{50\times 9.81}\\\Rightarrow h=4265.04\ \text{m}

Nancy could raise the weight to a maximum height of 4265.04\ \text{m}.

Mass of H_2 used per year = 25\times 10^{9}\ \text{kg/year}

Energy of H_2 = \dfrac{30\times10^9}{1000}=30\times 10^6\ \text{J/kg}

Power

P=25\times 10^{9}\ \text{kg/year}\times 30\times 10^6\ \text{J/kg}\\\Rightarrow P=7.5\times 10^{17}\ \text{J/year}\\\Rightarrow P=\dfrac{7.5\times 10^{17}}{365.25\times 24\times 60\times 60}\\\Rightarrow P=2.38\times 10^{10}\ \text{W}

The power requirement is 2.38\times 10^{10}\ \text{W}.

6 0
3 years ago
What are typical uses of Mainframe computer?​
madam [21]

a computer used primarily by large organizations for critical applications like bulk data processing for tasks such as censuses, industry and consumer statistics, enterprise resource planning, and large-scale transaction processing

7 0
3 years ago
An air conditioner operating at steady state maintains a dwelling at 70°F on a day when the outside temperature is 99°F. The rat
IrinaVladis [17]

Answer:

a) the coefficient of performance of the air conditioner is 3.5729

b)

- the power input required for a reversible air conditioner is 0.645 hp

- the coefficient of performance for the reversible air conditioner is 18.2759

Explanation:

Given the data in the question;

Lower Temperature T_L = 70°F = ( 70 + 460 )R = 530 R

Higher Temperature T_H = 99° F = ( 99 + 460 )R = 559 R

Cooling Load Q_L = 30000 Btu/h

we know that 1 hp = 2544.43 Btu/h

Net power input P = 3.3 hp = ( 3.3 × 2544.43 )Btu/h = 8396.619 Btu/h

a)

Coefficient of performance of the air conditioner;

COP_{air-condition = Cooling Load Q_L  / power P

we substitute

COP_{air-condition = 30000 Btu/h / 8396.619 Btu/h

COP_{air-condition = 3.5729

Therefore, the coefficient of performance of the air conditioner is 3.5729

b)

- Power input required ( in hp )

Q_L / P_{required = T_L / ( T_H - T_L )

we substitute

30000 Btu/h / P_{required = 530 R / ( 559 R - 530 R )

30000 Btu/h / P_{required = 530 R / 29 R

we solve for P_{required

P_{required  = ( 30000 Btu/h × 29 R ) / 530 R

P_{required  = ( 870000 Btu/h / 530 )

P_{required  = 1641.5094 Btu/h

we know that; 1 hp = 2544.43 Btu/h

so;

P_{required  = ( 1641.5094 / 2544.43 ) hp

P_{required  = 0.645 hp

Hence, the power input required for a reversible air conditioner is 0.645 hp

- the coefficient of performance for the reversible air conditioner;

COP_{rev-air-condition = T_L / ( T_H - T_L )

we substitute

COP_{rev-air-condition = 530 R / ( 559 R - 530 R )

COP_{rev-air-condition = 530 R / 29 R

COP_{rev-air-condition = 18.2759

Hence, the coefficient of performance for the reversible air conditioner is 18.2759

3 0
3 years ago
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