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3 years ago

8

What standard feature is on all circular saws?

1 answer:

PtichkaEL [24]3 years ago

4 0

What standard feature is on all circular saws?

4. All of the above

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A steel bar 110 mm long and having a square cross section 22 mm on an edge is pulled in tension with a load of 89,000 N, and exp

iragen [17]

Answer:

Elastic modulus of steel = 202.27 GPa

Explanation:

given data

long = 110 mm = 0.11 m

cross section 22 mm = 0.022 m

load = 89,000 N

elongation = 0.10 mm = 1 × $10^{-4}$ m

solution

we know that Elastic modulus is express as

Elastic modulus = $\frac{stress}{strain}$ ................1

here stress is

Stress = $\frac{Force}{area}$ .................2

Area = (0.022)²

and

Strain = $\frac{extension}{length}$ .............3

so here put value in equation 1 we get

Elastic modulus = $\frac{89,000\times 0.11}{0.022^2 \times 1 \times 10^{-4} }$

Elastic modulus of steel = 202.27 × $10^9$ Pa

Elastic modulus of steel = 202.27 GPa

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3 years ago

Show the ERD with relational notation with crowfoot. Your ERD must show PK, FKs, min and max cardinality, and correct line types

zhenek [66]

Answer

The answer and procedures of the exercise are attached in the following archives.

Step-by-step explanation:

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4 years ago

What flow type occurs when the heat exchanger has multiple passes and baffles?

Vanyuwa [196]

I think that the answer would be B or C

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3 years ago

Which examples best demonstrate likely tasks for Health, Safety, and Environmental Assurance workers? check all that apply

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Sam, Elijah, Joy Those are 100% correct from my human knowledge

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3 years ago

Read 2 more answers

Refrigerant-134a at 700 kPa, 70°C, and 7.2 kg/min is cooled by water in a condenser until it exists as a saturated liquid at the

alex41 [277]

Answer:

The mass flow rate of cooling water required to cool the refrigerant is $123.788\,\frac{kg}{min}$ .

Explanation:

A condenser is a heat exchanger used to cool working fluid (Refrigerant 134a) at the expense of cooling fluid (water), which works usually at steady state. Let suppose that there is no heat interactions between condenser and surroundings.The condenser is modelled after the First Law of Thermodynamics, which states:

$\dot Q_{ref} - \dot Q_{w} = 0$

$\dot Q_{ref} = \dot Q_{w}$

$\dot m_{ref}\cdot (h_{ref, in} - h_{ref,out}) = \dot m_{w}\cdot (h_{w, out} - h_{w,in})$

The mass flow rate of the cooling water is now cleared:

$\dot m_{w} = \dot m_{ref }\cdot \frac{h_{ref,in}-h_{ref,out}}{h_{w,out}-h_{w,in}}$

Given that $h_{ref,in} = 808.34\,\frac{kJ}{kg}$ , $h_{ref, out} = 88.82\,\frac{kJ}{kg}$ , $h_{w,out} = 104.83\,\frac{kJ}{kg}$ and $h_{w,in} = 62.98\,\frac{kJ}{kg}$ , the mass flow of the cooling water is:

$\dot m_{w} = \left(7.2\,\frac{kg}{min} \right)\cdot \left(\frac{808.34\,\frac{kJ}{kg}-88.82\,\frac{kJ}{kg} }{104.83\,\frac{kJ}{kg}-62.98\,\frac{kJ}{kg} } \right)$

$\dot m_{w} = 123.788\,\frac{kg}{min}$

The mass flow rate of cooling water required to cool the refrigerant is $123.788\,\frac{kg}{min}$ .

4 0

3 years ago