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3 years ago

When must an Assured Equipment Grounding Conductor Program (AEGCP) be in place?

Engineering

1 answer:

sasho [114]3 years ago

8 0

Answer:

Only when working around power lines

Explanation:

Assured Equipment Grounding Conductor Program (AEGCP) is applicable mainly in sites dealing with power. Therefore, the most appropriate worksite to have Assured Equipment Grounding Conductor Program is when working around power lines.

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Why is the face of the claw on a claw hammer usually a smooth curve? Why isn't it straight or some other shape?

GarryVolchara [31]

Answer:

The face of the claw on the claw hammer is usually a smooth curve so as to improve the ease with which nails are removed when removing nails because as the nail held between the V shaped split claw is being pulled out from the wood, it slides more and more towards cheek, reducing the distance of the nail from the cheek which is the fulcrum, thereby increasing the mechanical advantage because the location of the hand on the grip remains unchanged

Explanation:

7 0

3 years ago

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

1. The area(in square centimeters) of a square coaster can be represented by

Brrunno [24]

Answer: (d−4) 2

Explanation: Factoring d2-8d+16

The first term is, d2 its coefficient is 1 .

The middle term is, -8d its coefficient is -8 .

The last term, "the constant", is +16

Step-1 : Multiply the coefficient of the first term by the constant 1 • 16 = 16

Step-2 : Find two factors of 16 whose sum equals the coefficient of the middle term, which is -8 .

-16 + -1 = -17

-8 + -2 = -10

-4 + -4 = -8 That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -4 and -4

d2 - 4d - 4d - 16

Step-4 : Add up the first 2 terms, pulling out like factors :

d • (d-4)

Add up the last 2 terms, pulling out common factors :

4 • (d-4)

Step-5 : Add up the four terms of step 4 :

(d-4) • (d-4)

Which is the desired factorization

Multiply (d-4) by (d-4)

The rule says : To multiply exponential expressions which have the same base, add up their exponents.

In our case, the common base is (d-4) and the exponents are :

1 , as (d-4) is the same number as (d-4)1

and 1 , as (d-4) is the same number as (d-4)1

The product is therefore, (d-4)(1+1) = (d-4)2

Please mark me brainlyest

7 0

3 years ago

The horizontal component of acceleration, ay during a projectile motion is usually assumed to be_________ a)-9.81 m/s^2 b)-Zero

Alenkinab [10]

Answer:

b) zero

Explanation:

The horizontal component of acceleration during projectile motion is usually assumed to be zero.Because in projectile motion horizontal component of velocity will remain the constant and we know that rate of change of velocity with time is called acceleration.So when velocity is constant then acceleration will be zero.

In projectile motion ,gravitational acceleration will be in only vertical direction.

6 0

3 years ago

The mechanical properties of a metal may be improved by incorporating fine particles of its oxide. Given that the moduli of elas

Lilit [14]

Answer:

A) 209.12 GPa

B) 105.41 GPa

Explanation:

We are given;

Modulus of elasticity of the metal; E_m = 67 GPa

Modulus of elasticity of the oxide; E_f = 390 GPa

Composition of oxide particles; V_f = 44% = 0.44

A) Formula for upper bound modulus of elasticity is given as;

E = E_m(1 - V_f) + (E_f × V_f)

Plugging in the relevant values gives;

E = (67(1 - 0.44)) + (390 × 0.44)

E = 209.12 GPa

B) Formula for upper bound modulus of elasticity is given as;

E = 1/[(V_f/E_f) + (1 - V_f)/E_m]

Plugging in the relevant values;

E = 1/((0.44/390) + ((1 - 0.44)/67))

E = 105.41 GPa

4 0

2 years ago