I have to invent something for people 65pluse so I need to know what are something old people would bye.

To prevent drainage of the transmission fluid from the converter when the

-BARSIC- [3]

Answer
D I think
Explanation

4 0

3 years ago

An equal-tangent sag vertical curve (with a negative initial and a positive final grade) is designed for 55 mi/h. The PVI is at

Varvara68 [4.7K]

Answer:

The lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.

Explanation:

Length of curve is given as

$L=2(PVT-PVI)\\L=2(242+30-240+00)\\L=2(230)\\L=460 ft$

$G_2$ is given as

$G_2=\frac{E_{PVT}-E_{PVI}}{0.5L}\\G_2=\frac{127.5-122}{0.5*460}\\G_2=0.025=2.5 \%$

The K value is given from the table 3.3 for 55 mi/hr is 115. So the value of A is given as

$A=\frac{L}{K}\\A=\frac{460}{115}\\A=4$

A is given as

$-G_1=A-G_2\\-G_1=4.0-2.5\\-G_1=1.5\\G_1=-1.5\%$

With initial grade, the elevation of PVC is

$E_{PVC}=E_{PVI}+G_1(L/2)\\E_{PVC}=122+1.5%(460/2)\\E_{PVC}=125.45 ft\\$

The station is given as

$St_{PVC}=St_{PVI}-(L/2)\\St_{PVC}=24000-(230)\\St_{PVC}=237+70\\$

Low point is given as

$x=K \times |G_1|\\x=115 \times 1.5\\x=172.5 ft$

The station of low point is given as

$St_{low}=St_{PVC}-(x)\\St_{low}=23770+(172.5)\\St_{low}=239+42.5 ft\\$

The elevation is given as

$E_{low}=\frac{G_2-G_1}{2L} x^2+G_1x+E_{PVC}\\E_{low}=\frac{2.5-(-1.5)}{2*460} (1.72)^2+(-1.5)*(1.72)+125.45\\E_{low}=124.16 ft$

So the lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.

3 0

4 years ago

Scenario* The organization is a regional XYZ Credit Union/Bank that has multiple branches and locations throughout the region.*

fiasKO [112]

Answer:

1) For preparation of plan :

It is to gather the team and creating the timeline. Gathering the inputs for the plan and confirming the mission and vision then launching the strategy.

2) A mission would describe about the company and the specification of the company

A vision is the long term or future state of the organization to achieve in a time frame.

Both would include the OAS statements and strategic shifts.

3) The OAS and strategic shift would help to create a balanced score card that would help for data gathering and analysis

This would include the financial, customers goal, Process goals, people's goals etc.

4) To create and develop the plan's framework:

To set the objective for high level organizational goals

To measure the and accomplish the objectives

To take initiatives to achieve the objectives

Explanation:

8 0

3 years ago

Match the word with the definition:

aksik [14]

1. Renewable Resources = (Renewable means you can keep making it) = resources that can be replenished (such as trees)

2. Nonrenewable Resources = ( Nonrenewable means it can't be made once it is used up) = resources that are gone once they are used (such as fossil fuels)

3. Producer = ( produces something) = person who makes goods or provides services

4. Consumer = ( uses something) = person whose wants are satisfied by using goods and services

5. Allocate = ( put someplace) = distribute

6. Choice = option

7 0

3 years ago

Read 2 more answers

A 4-L pressure cooker has an operating pressure of 175 kPa. Initially, one-half of the volume is filled with liquid and the othe

vodomira [7]

Answer:

the highest rate of heat transfer allowed is 0.9306 kW

Explanation:

Given the data in the question;

Volume = 4L = 0.004 m³

V $_f$ = V $_g$ = 0.002 m³

Using Table ( saturated water - pressure table);

at pressure p = 175 kPa;

v $_f$ = 0.001057 m³/kg

v $_g$ = 1.0037 m³/kg

u $_f$ = 486.82 kJ/kg

u $_g$ 2524.5 kJ/kg

h $_g$ = 2700.2 kJ/kg

So the initial mass of the water;

m₁ = V $_f$ /v $_f$ + V $_g$ /v $_g$

we substitute

m₁ = 0.002/0.001057 + 0.002/1.0037

m₁ = 1.89414 kg

Now, the final mass will be;

m₂ = V/v $_g$

m₂ = 0.004 / 1.0037

m₂ = 0.003985 kg

Now, mass leaving the pressure cooker is;

m $_{out$ = m₁ - m₂

m $_{out$ = 1.89414 - 0.003985

m $_{out$ = 1.890155 kg

so, Initial internal energy will be;

U₁ = m $_f$ u $_f$ + m $_g$ u $_g$

U₁ = (V $_f$ /v $_f$ )u $_f$ + (V $_g$ /v $_g$ )u $_g$

we substitute

U₁ = (0.002/0.001057)(486.82) + (0.002/1.0037)(2524.5)

U₁ = 921.135288 + 5.030387

U₁ = 926.165675 kJ

Now, using Energy balance;

E $_{in$ - E $_{out$ = ΔE $_{sys$

QΔt - m $_{out$ h $_{out$ = m₂u₂ - U₁

QΔt - m $_{out$ h $_g$ = m₂u $_g$ - U₁

given that time = 75 min = 75 × 60s = 4500 sec

so we substitute

Q(4500) - ( 1.890155 × 2700.2 ) = ( 0.003985 × 2524.5 ) - 926.165675

Q(4500) - 5103.7965 = 10.06013 - 926.165675

Q(4500) = 10.06013 - 926.165675 + 5103.7965

Q(4500) = 4187.690955

Q = 4187.690955 / 4500

Q = 0.9306 kW

Therefore, the highest rate of heat transfer allowed is 0.9306 kW

5 0

3 years ago