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2 years ago

14

“In a trusting relationship, confidential information is kept confidential.” Explain what the limits to confidentiality are and

give an example.

1 answer:

Sliva [168]2 years ago

7 0

The limits are dangerous situation. Ex. Suicidal actions/thoughts, anything that may put yourself, the individual, or anyone else at risk.

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Guyss I seriously and urgently need help what are the steps to build a headgear ??

beks73 [17]

Answer:

Design. Modern headframes are built out of steel, concrete or a combination of both. Timber headframes are no longer used in industrialized countries, but are still used in developing nations

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2 years ago

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In the planning process of the product development life cycle what is it important to inventory

Verizon [17]

Your Answer would be A I believe.

6 0

2 years ago

A four-cylinder, four-stroke internal combustion engine has a bore of 3.7 in. and a stroke of 3.4 in. The clearance volume is 16

abruzzese [7]

Answer:

1) The three possible assumptions are

a) All processes are reversible internally

b) Air, which is the working fluid circulates continuously in a closed loop

cycle

c) The process of combustion is depicted as a heat addition process

2) The diagrams are attached

5) The net work per cycle is 845.88 kJ/kg

The power developed in horsepower ≈ 45374 hP

Explanation:

1) The three possible assumptions are

a) All processes are reversible internally

b) Air, which is the working fluid circulates continuously in a closed loop

cycle

c) The process of combustion is depicted as a heat addition process

2) The diagrams are attached

5) The dimension of the cylinder bore diameter = 3.7 in. = 0.09398 m

Stroke length = 3.4 in. = 0.08636 m.

The volume of the cylinder v₁= 0.08636 ×(0.09398²)/4 = 5.99×10⁻⁴ m³

The clearance volume = 16% of cylinder volume = 0.16×5.99×10⁻⁴ m³

The clearance volume, v₂ = 9.59 × 10⁻⁵ m³

p₁ = 14.5 lbf/in.² = 99973.981 Pa

T₁ = 60 F = 288.706 K

$\dfrac{T_{2}}{T_{1}} = \left (\dfrac{v_{1}}{v_{2}} \right )^{K-1}$

Otto cycle T-S diagram

T₂ = 288.706* $6.25^{0.393}$ = 592.984 K

The maximum temperature = T₃ = 5200 R = 2888.89 K

$\dfrac{T_{3}}{T_{4}} = \left (\dfrac{v_{4}}{v_{3}} \right )^{K-1}$

T₄ = 2888.89 / $6.25^{0.393}$ = 1406.5 K

Work done, W = $c_v$ ×(T₃ - T₂) - $c_v$ ×(T₄ - T₁)

0.718×(2888.89 - 592.984) - 0.718×(1406.5 - 288.706) = 845.88 kJ/kg

The power developed in an Otto cycle = W×Cycle per second

= 845.88 × 2400 / 60 = 33,835.377 kW = 45373.99 ≈ 45374 hP.

8 0

3 years ago

For a steel alloy it has been determined that a carburizing heat treatment of 15 h duration will raise the carbon concentration

Amiraneli [1.4K]

Answer:

135 hour

Explanation:

It is given that a carburizing heat treatment of 15 hour will raise the carbon concentration by 0.35 wt% at a point of 2 mm from the surface.

We have to find the time necessary to achieve the same concentration at a 6 mm position.

we know that $\frac{x_1^2}{Dt}=constant$ where x is distance and t is time .As the temperature is constant so D will be also constant

So $\frac{x_1^2}{t}=constant$

then $\frac{x_1^2}{t_1}=\frac{x_2^2}{t_2}$ we have given $x_1=2 mm\ ,t_1=15 hour\ ,x_2=6\ mm$ and we have to find $t_2$ putting all these value in equation

$\frac{2^2}{15}=\frac{6^2}{t_2}$

so $t_2=135\ hour$

5 0

3 years ago

"At 195 miles long, and with 7,325 miles of coastline, the Chesapeake Bay is the largest and most complex estuary in the United

Paraphin [41]

Answer:

see explaination

Explanation:

Part a) Width of bay at Potomac River:

Given Data:

· Actual Width at Potomac River = 30 miles

· Bay Model Length Ratio Lr = 1/1000

In fluid mechanics models of real structures are prepared in simulation so that they can be analyzed accurately. A model is known to have simulation if model carries same geometric, kinematic and dynamic properties at a small scale.

Length of any part in model = Actual length x Lr

Hence,

Model Width of bay at Potomac River = 30 x 1/1000 = 0.03 miles

Since 1 mile = 5280 ft

Model Width of bay at Potomac River = 0.03 x 5280 = 158.4 ft

Part b) Model Length of bay bridge in model:

Given Data:

· Actual Length of bay bridge = 4.3 miles

· Bay Model Length Ratio Lr = 1/1000

Model Length = Actual Length x Lr = 4.3 x 1/1000 = 0.0043 miles

Since 1 mile = 5280 ft

Model Length in feet = 0.0043 x 5280 = 22.704 ft

Part c) Model Length of bay bridge in model:

Given Data:

· Model Area = 8 acre

· Bay Model Length Ratio Lr = 1/1000

Model Area = Actual Area x Lr x Lr

8 Model Area :: Actual Area =- (Lp)2 2 = 8,000,000 acre 1000

Since 1 square mile = 640 acre,

Actual Area in square miles = 8,000,000/640 = 12,500 square miles

Part d) Average and maximum depth of model:

Given Data:

· Actual Average depth = 28 ft

· Actual Maximum depth = 174 ft

· Bay Model Length Ratio Lr = 1/1000

Model average depth = Actual average depth x Lr = 28 x 1/1000 = 0.028 feet

Since 1 ft = 12 inch

Model average depth in inch = 0.028 x 12 = 0.336 in

Model maximum depth = Actual maximum depth x Lr = 174 x 1/1000 = 0.174 feet

Since 1 ft = 12 inch

Model maximum depth in inch = 0.174 x 12 = 2.088 in

4 0

3 years ago