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Rina8888 [55]
3 years ago
13

Consider the following example: The 28-day compressive strength should be 4,000 psi. The slump should be between 3 and 4 in. and

the maximum aggregate size should not exceed 1 in. The coarse and fine aggregates in the storage bins are wet. The properties of the materials are as follows:________.
Cement : Type I, specific gravity = 3.15
Coarse Aggregate: Bulk specific gravity (SSD) = 2.70; absorption
capacity = 1.1%; dry-rodded unit weight = 105 lb./ft.3
surface moisture = 1%
Fine Aggregate: Bulk specific gravity (SSD) = 2.67; absorption
capacity = 1.3%; fineness modulus = 2.70;
surface moisture = 1.5%
Engineering
1 answer:
lara [203]3 years ago
8 0
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Technician A says that hoods are designed with reinforcements to prevent folding during a collision. Technician B says that some
-BARSIC- [3]

Technician A is wrong.

  • Usually, hoods have what is called "Crush Zones" underneath the panels. The function of the Crush Zone is to prevent the hoods, during a collision, from entering into the passenger space.

  • The crush zones allow the hoods to fold instead.

Technician B is right.

  • Automobile producers now make use of a hybrid form of hood that consists of fiberglass reinforced with plastic.

  • They are mostly used for trucks that have a low volume of production.

  • The hood is built using a process called Resin Transfer Model (RTM).

See the link below for more about automobile engineering:

brainly.com/question/4822721

6 0
2 years ago
Risk Management in a Business ModelLearning Objectives and OutcomesCreate a report documenting various aspects of how risk mana
zubka84 [21]

Answer:

The question is explained in detailed way in explanation section and in attached files.

Explanation:

The HIPAA Security Rule is designed to be flexible and appropriate for our organization’s particular size, structure, and inherent risks to business and personal information. Risk analysis is meant to be an ongoing process, during which we regularly review our records to track access to business and personal systems and data. With this in mind, I recommend that we expand our information security strategy to include more than just what is required in HIPAA. Just as a reminder below is the HIPAA ecompliance and implementation strategy that we came up with last week as given in attached file 1.

There are several areas in IT security that the above is incomplete or insufficient in. We recommend implementing several more complete or alternative controls in order to protect our systems, patients, employees, contractors, vendors, and assets beyond the HIPAA minimum requirements. The below section describes what the Centers for Medicare and Medicaid Services (CMS) recommend as additional areas to focus on in the effort to increase an organization's security. (See the attached file # example of some of the areas that we should monitor beyond what HIPAA requires are given in attached file # 03.

3 0
3 years ago
During an expansion process, the pressure of a gas changes from 15 to 140 psia according to the relation P = aV + b, where a = 5
Dominik [7]

The work done during the process is 359 btu

<u></u>

<u>Explanation:</u>

Given-

P1 = 15psia

P2 = 140 psia

V1 = 7ft³

a = 5 psia/ft³

b = C

P = aV +b

Work done, W = ?

P1 = aV1 + b

15 = 5 (7) + b

b = -20 psia

P2 = aV2 + b

140 = 5 ( V2) - 20

V2 = 32 ft³

The work done by the process is the area under the curve which is trapezoidal.

Therefore,

Work done, W = area of trapezoid

= (P2 + P1 / 2) (V2 - V1)

= ( 140 + 15 / 2 ) ( 32 - 7)

= 1937.5 psia ft³

= 1937.5/ 5.4039 = 359 btu

Therefore, the work done during the process is 359 btu

5 0
3 years ago
Assume the average fuel flow rate at the peak torque speed (1500 rpm) is 15kg/hr for a sixcylinder four-stroke diesel engine und
sveticcg [70]

Answer:

Q = 8.845 DEGREE

Explanation:

given data:

combine Mass for 6 cylinder (M) =15 Kg/hr

mass of  each cylinder (m) = 15/6 = 2.5 Kg/hr = 0.000694 Kg/ sec

Engine speed (N)= 1500rpm

Diameter of one nozzle hole ( d) = 200 micrometer = 0.0002 m

Discharge Coefficient (Cd) = 0.75

Pressure difference = 100 MPa

Density of fuel = 800 kg/m^3

velocity of fuel is v  = cd\sqrt{\frac{2*P}{p}}

v = 0.75 \sqrt{\frac{2\times 100\times 10^6}{800}} = 375 m/sec

injected fuel volume  (V) =Area of given  Orifices × Fuel velocity × time of single injection × no of injection/sec

we know that p = m/ V

SoV = \frac{0.000694}{800} =8.68\times10^{-7} m3/sec

putting these value in volume equation and solve for Discharge 8.68\times 10^{-7} = (\frac{(3.14}{4})\times 6\times( .0002\times .0002) \times  375 \times  \frac{(Q}{360}) \times \frac{30}{750} \times \frac{(750}{60)}

Q = 8.845 DEGREE

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3 years ago
The artifacts that expand what humans are able to do are called:.?
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