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Naddik [55]
3 years ago
12

Lobbyists are hired by companies to influence legislation to meet a company's goals. They ask for things needed to program plans

related to health care and come together as _____.
a.political pack groups
b. professional and amateur groups
c. social media groups
d. special interest groups
Engineering
2 answers:
denis-greek [22]3 years ago
8 0
Social media groups. So c
goldenfox [79]3 years ago
3 0
The answer is c ………udieiofnfe
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An electric water heater held at 120° F is kept in a 60°F room. When purchased, its insulation is equivalent to R-5. An owner pu
Trava [24]

Answer:

Total saving would be of 36.917 $\yr

Explanation:

Given Data:

T_{heater} = 120 Degree F

T_{room} = 60 Degree F

A = 30 ft^2

\eta = 100%

Heat loss before previous final value = \frac{A \Delta T}{R}

                                                              =\frac{30\times *(120-60)}{5}

                                                              = 360 Btu/hr

Heat loss after new value= \frac{30\times \times (120-60)}{15} = 120 Btu/hr

saving would be = 360 - 120 Btu/hr \times kw hr/ 3412 Btu\times 24 hr/day \times 365 day/year

                           = 616.1782 kw hr/yr

cost = 616.1782 \times 0.06$

         = 36.917 $\yr

6 0
4 years ago
Suppose that a wing component on an aircraft is fabricated from an aluminum alloy that has a plane-strain fracture toughness of
Orlov [11]

Answer: 133.88 MPa approximately 134 MPa

Explanation:

Given

Plane strains fracture toughness, k = 26 MPa

Stress at which fracture occurs, σ = 112 MPa

Maximum internal crack length, l = 8.6 mm = 8.6*10^-3 m

Critical internal crack length, l' = 6 mm = 6*10^-3 m

We know that

σ = K/(Y.√πa), where

112 MPa = 26 MPa / Y.√[3.142 * 8.6*10^-3)/2]

112 MPa = 26 MPa / Y.√(3.142 * 0.043)

112 = 26 / Y.√1.35*10^-2

112 = 26 / Y * 0.116

Y = 26 / 112 * 0.116

Y = 26 / 13

Y = 2

σ = K/(Y.√πa), using l'instead of l and, using Y as 2

σ = 26 / 2 * [√3.142 * (6*10^-3/2)]

σ = 26 / 2 * √(3.142 *3*10^-3)

σ = 26 / 2 * √0.009426

σ = 26 / 2 * 0.0971

σ = 26 / 0.1942

σ = 133.88 MPa

8 0
3 years ago
What chemicals go into lake michigan
solniwko [45]

Answer:

They include PCBs, dioxins, furans, hexachlorobenzene, benzo(a)pyrene (one of many polyaromatic hydrocarbons or PAHs), lead and mercury compounds and pesticides (toxaphene, mirex, aldrin/dieldrin, DDT). 3 Great Lakes Water Quality Agreement Work Group Report on Chemicals of Emerging Concern, August 2009.

7 0
4 years ago
Read 2 more answers
Food, animals, wood and alcohol are examples of:
Fantom [35]
I pretty sure it would be A
4 0
3 years ago
Read 2 more answers
One kilogram of water fills a 0.140 m^3 rigid container at an initial pressure of 1.8 MPa. The container is then cooled to 40°C.
Pavel [41]

Answer:

T1 = 299.18 °C

P2 = 0.00738443 MPa

Explanation:

From the data, we can get two properties for the initial condition. These are pressure and specific volume.

The pressure is 1.8 MPa and the specific volume, we can get it with the mass and volume of the container, since it’s filled this is also the volume of the water in it.

v=\frac{vol (m^{3})}{mass (kg)} = \frac{0.140 m^{3}}{1 kg} = 0.140 \frac{ m^{3}}{kg}

When we check in the thermodynamic tables, the conditions for saturation at 1.8 MPa we found the following:

P^{sat} = 1.8 MPa

T^{sat} = 207.12 C

v_{g} = 0.1103\frac {m^{3}}{kg} specific volume for the saturated vapor  

v_{l} = 0.001167 \frac{m^{3}}{kg} specific volume for the saturated liquid  

Since the specific volume in our condition is higher that the specific volume for the saturated vapor, we have a superheated steam.  

Looking in the thermodynamic tables for superheated steam we found that the temperature where the steam has a specific volume of 0.140 \frac{ m^{3}}{kg} at 1.8 MPa is 299.18 °C. This is the initial temperature in the container.

Since the only information that we have about the final condition is that the container was cooled. We can assume that it was cooled until a condition of saturation. So, the final pressure for the water will be the pressure of saturation for a temperature of 40°C. From thermodynamic tables we get:

P^{sat} at 40C = 0.00738443 MPa

7 0
3 years ago
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