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dangina [55]
3 years ago
12

How much will it cost to train the entire company to use a recycling program if the training includes paper handouts? (Remember,

the training will take one hour and a team of three people will take three hours each to create the training program.) With the current budget, is it possible to train the company and buy recycling carts if recycling pickup in your city is free? Assume the best choice for recycle carts in your area was $80 per cart.
Engineering
1 answer:
bezimeni [28]3 years ago
8 0

Answer:

With the current budget of $1,000, it is possible to train the company and buy 3 recycling carts given that recycling pickup within the area is free

Explanation:

The office waste management budget that is not being spent = $1,000

The given expense parameters are;

The training cost per hour per employee = $12.00

The number of people in the entire company = 45

The cost of preparing the training material per hour = $20.00

The number of people to create the training material = 3 people

The time it will take each person in creating the training material = 3 hours

The cost of recycling cart = $80.00 per cart

The cost of paper handouts per employee = $0.05

The cost of materials include;

The total cost for the training = $12.00 × 45 = $540

The cost for the handout = $0.05 × 45 = $2.25

The cost of preparing the materials = $20.00 × 3 × 3 = $180.00

The total costs of the training = $540 + $4 + $180.00 = $724

The amount available to buy cart = $1,000 - $724 = $276

Therefore;

The amount available to buy cart = $276

The number of carts that can be bought = 276/80 = 3.45 carts

Therefore, we round down to get;

The number of carts that can be bought = 3 carts

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A rotating shaft is subjected to a steady torsional stress of 13 ksi and an alternating bending stress of 22 ksi.
mixas84 [53]

Answer:

A) б1 = 28 ksi and  б2 = -6.02 ksi

B) 1.25

Explanation:

Given data :

Torsional stress = 13 ksi

Alternating bending stress = 22ksi

A) determine yielding factor of safety  according to the distortion energy theory

б1,2 = \frac{22}{2} ± √(22/2)² + 13²

       = 11  ± 17

therefore б1 = 28 ksi  hence б2 = -6.02 ksi

B) determine the fatigue factor of safety  

with properties ;  Se = 35ksi, Sy = 60 ksi, Sut = 85 ksi

( б1 - б2 )²  + ( б2 - б3 )² + ( б3 - б1 )²  ≤  2 ( Sy / FOS ) ²

( 28 + 6.02 ) ² + ( 6.02 - 0 )² + ( 0 - 28 )² ≤  2 ( 60 / FOS ) ²

solving for FOS = 1.9

Next we can determine FOS with the use of Goodman criterion

бm / Sut  + бa / Se  =  1 / FOS

= 0 / 85 + 28/35 = 1 / FOS

making FOS the subject of the equation ; hence  FOS = 1.25

3 0
3 years ago
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4 0
3 years ago
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3 years ago
For some alloy, the yield stress is 345-MPa (50,000-psi) and the elastic modulus (E) is 103-GPa (15x106 psi). What is the maximu
OverLord2011 [107]

The maximum load that may be applied to a specimen with a cross-sectional area of 130 mm² is; 35535 N

<h3>How to find Elastic Modulus?</h3>

We are told that for an alloy, the yield stress is 345-MPa and the elastic modulus (E) is 103-GPa.

Now, we want to find the maximum load that may be applied to a specimen with a cross-sectional area of 130-mm² without plastic deformation. Thus;

We are given the parameters;

Yield Stress; σ = 345 Mpa = 345 * 10⁶ Pa

Elastic Modulus; E = 103 GPa = 103 * 10⁹ Pa

Cross sectional Area; A = 130 mm² = 103 * 10⁻⁶ m²

Formula for stress without Plastic deformation is;

σ = F_max/Area

where;

σ is stress

F_max is maximum force

Area is Area

Thus making maximum force the subject of the formula gives;

F_max = σ * A

Plugging in the relevant values for stress and area gives us;

F_max = 345 * 10⁶ * 103 * 10⁻⁶

F_max = 35535 N

The maximum load that may be applied to a specimen with a cross-sectional area of 130 mm² is gotten to be 35535 N

Read more about Elastic Modulus at; brainly.com/question/6864866

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4 0
2 years ago
True or false: The maximum tensile force a solid, cylindrical wire can withstand increases as the thickness of the wire increase
oee [108]

Answer: I believe it's true

Explanation: Because for instance if you keep adding wire around it'self it will increase in thickness, which in any case it would also increase in it durability.

7 0
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