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Kobotan [32]
2 years ago
4

The cost of electricity for running production equipment is classified as: Conversion cost Period cost A) Yes No B) Yes Yes C) N

o Yes D) No No
Engineering
2 answers:
Y_Kistochka [10]2 years ago
7 0

A) Yes No

Explanation:

Conversion costs are those costs that combine both labor costs and manufacturing overhead costs. These are those that affect the final cost of the manufactured product and the production costs.These costs are a mandatory in the conversion of the raw material to final product. In this case, electricity for running the production equipment effects final cost of the product produced.Period costs are not related to production of a product.For example marketing costs.They are recorded in the income statement as expenses on the income.

Learn More

Conversion costs :brainly.com/question/14347362

Period costs:brainly.com/question/14265348

Keywords : cost. electricity, production equipment,

#LearnwithBrainly

allsm [11]2 years ago
6 0

Your answer should be:

A) Yes No

plz mark brainliest

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Milton has been tracking the migrating patterns of whales in the northwest Atlantic Ocean for five years. He knows where and whe
vovangra [49]

Answer:

knowledge of animal behavior and anatomy

Explanation:

the qualification that will make Milton successful in his research is a knowledge of animal behaviour and also their anatomy. the knowledge of whales behaviour has opened his eyes into their world so he knows to a great deal about them. it is through his knowledge of the behaviour of whales that he's able to get used to their migrating patterns to know where and when to find them. Also, through the body anatomy of whales he knows what their movement is like.

8 0
2 years ago
Read 2 more answers
You are working in a lab where RC circuits are used to delay the initiation of a process. One particular experiment involves an
Ymorist [56]

Answer:

t'_{1\2} = 6.6 sec

Explanation:

the half life of the given circuit is given by

t_{1\2} =\tau ln2

where [/tex]\tau = RC[/tex]

t_{1\2} = RCln2

Given t_{1\2} = 3 sec

resistance in the circuit is 40 ohm and to extend the half cycle we added new resister of 48 ohm. the net resitance is 40+48 = 88 ohms

now the new half life is

t'_{1\2} =R'Cln2

Divide equation 2 by 1

\frac{t'_{1\2}}{t_{1\2}} = \frac{R'Cln2}{RCln2} = \frac{R'}{R}

t'_{1\2} = t'_{1\2}\frac{R'}{R}

putting all value we get new half life

t'_{1\2} = 3 * \frac{88}{40}  = 6.6 sec

t'_{1\2} = 6.6 sec

7 0
2 years ago
The u velocity component of a steady, two-dimensional, incompressible flow field is u = 3 ax 2 - 2 bxy, where a and b are consta
Kruka [31]

Answer:

The velocity component v is -6axy+2by^2+f(x)

Explanation:

Given that,

The velocity component of a steady, two-dimensional

u=3ax^2-2bxy

We need to calculate the function of x

Using given equation

u=3ax^2-2bxy

Where, a and b is constant

On differential

\dfrac{du}{dx}=6ax-2by

We need to calculate the velocity component v

Using equation of velocity

\dfrac{dv}{dy}=-\dfrac{du}{dx}-\dfrac{dw}{dz}

Put the value into the formula

\dfrac{dv}{dy}=-6ax+2by-0

Now, on integration w.r.t y

v=-6axy+2by^2+f(x)

Hence, The velocity component v is -6axy+2by^2+f(x)

4 0
2 years ago
Define a) Principal Plane b) Principal Stress c) anelasticity d) yield point e) ultimate tensile stress f) hardness g) toughness
bagirrra123 [75]

Answer:

Principal Plane: It is that plane in a stressed body over which no shearing stresses act. As we know that in a stressed body on different planes 2 different kind of stresses act normal stresses acting normal to the plane ans shearing stresses acting in the plane. The special planes over which no shearing stresses act and only normal stresses are present are termed as principal planes.

Principal Stress: The stresses in the principal planes are termed as normal stresses.

Anelasticity: It is the behavior of a material in which no definite relation can found to exist between stress and strain at any point in the stressed body.

Yield Point: It is the point in the stress-strain curve of a body at which the stress in the body reaches it's yield value or the object is just about to undergo plastic deformation if we just increase value of stress above this value. It is often not well defined in high strength materials or in some materials such as mild steel 2 yield points are observed.

Ultimate tensile strength: It is the maximum value of stress that a body can develop prior to fracture.

Hardness: it is defined as the ability of the body to resist scratches or indentation or abrasion.

Toughness: It is the ability of the body to absorb energy and deform without fracture when it is loaded. The area under the stress strain curve is taken as a measure of toughness of the body.

Elastic limit: The stress limit upto  which the body regains it's original shape upon removal of the stresses is termed as elastic limit of the body.

4 0
2 years ago
Link BD consists of a single bar 36 mm wide and 18 mm thick. Knowing that each pin has a 12-mm diameter, determine the maximum v
MAXImum [283]

Answer:

hello the diagram attached to your question is missing attached below is the missing diagram

answer :

a) 48.11 MPa

b) - 55.55 MPa

Explanation:

First we consider the equilibrium moments about point A

∑ Ma = 0

( Fbd * 300cos30° ) + ( 24sin∅ * 450cos30° ) - ( 24cos∅ * 450sin30° ) = 0

therefore ;<em> Fbd = 36 ( cos ∅tan30° - sin∅ ) kN  ----- ( 1 )</em>

A ) when ∅ = 0

Fbd = 20.7846 kN

link BD will be under tension when ∅ = 0, hence we will calculate the loading area using this equation

A = ( b - d ) t

b = 12 mm

d = 36 mm

t = 18

therefore loading area ( A ) = 432 mm^2

determine the maximum value of average normal stress in link BD  using the relation below

бbd = \frac{Fbd}{A}  = 20.7846 kN / 432 mm^2  =  48.11 MPa

b) when ∅ = 90°

Fbd = -36 kN

the negativity indicate that the loading direction is in contrast to the assumed direction of loading

There is compression in link BD

next we have to calculate the loading area using this equation ;

A = b * t

b = 36mm

t = 18mm

hence loading area = 36 * 18 = 648 mm^2

determine the maximum value of average normal stress in link BD  using the relation below

бbd = \frac{Fbd}{A} = -36 kN / 648mm^2 = -55.55 MPa

4 0
1 year ago
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