What was a campaign belief in the 1980 presidential election? Carter called for a stronger national defense. Carter promised to appoint a woman to the Supreme Court. Reagan endorsed a conservative economic policy. Reagan supported the Equal Rights Amendment.

Answers

Answer 1

Answer:

C-Reagan endorsed a conservative economic policy.

Explanation:

edge21

Answer 2

Reagan endorsed a conservative economic policy.

What is Presidential election?

In the US, presidential elections take place every four years. The next election for the nation's top position will be in 2024 because the last one occurred in 2020.

Federal elections must take place "the Tuesday next after the first Monday in November," according to U.S. law code.

Every even-numbered year has a presidential election and a congressional midterm election (e.g. 2016, 2018, 2020.)

Therefore, Reagan endorsed a conservative economic policy.

To learn more about economic policy, refer to the link:

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Related Questions

A pointer is spun on a fair wheel of chance having its periphery labeled Trom 0 to 100. (a) Whhat is the sample space for this experiment? (b)What is the probability that the pointer will stop between 20 and 35? (c) What is the probability that the wheel will stop on 58?

Answers

Answer:

A pointer is spun on a fair wheel of chance having its periphery labeled Trom 0 to 100. (a) Whhat is the sample space for this experiment? (b)What is the probability that the pointer will stop between 20 and 35? (c) What is the probability that the wheel will stop on 58?

Explanation:

thats all you said

Answer:

hii my name is RAGHAV what is your name

Explanation:

this question is which chapter

You have a 20 Volt power source attached to a light bulb that you've measured has a resistance of 8 Ohms, what is the power output of this light bulb (in Watts)?

Answers

Answer:

50 Watts

Explanation:

For this problem, we simply apply Ohm's law:

V = IR

V / R = I

P = IV

P = ( V / R ) V

P = ( V^2 / R )

P = (20 V)^2 / 8 Ω

P = 400 V^2 / 8 Ω

P = 50 Watts

Hence, the power consumed by the lightbulb is 50 Watts.

Cheers.

When water levels on the exterior of a building exceed water levels on the interior, hydrostatic loads become:_______

Answers

Answer: more dense

Explanation:

1:
Determine the dynamic pressure of water received at the site described below.



Water Tower holds water at an elevation of 265 feet

Site is at an elevation of 145 feet

The water supply system uses cast iron pipes.

The water travels through 3.2 miles of pipes before reaching the site

The pipe has a diameter of 8 inches

The water travels through 9 90-degree bends, 4 Branch tees, and 1 Swing Check Valve

The water has a flow rate of 105 gpm



Round to the hundredths place (2 places after the decimal)




2:
What is the static head of a water supply system if the water tower holds water at and elevation of 462 feet and the site that uses the water is at an elevation of 294 feet?
Answer value

Answers

Answer:

20.87 Pa

Explanation:

The formula for dynamic pressure is given as;

q= 1/2*ρ*v²

where ;

q=dynamic pressure

ρ = density of fluid

v = velocity of fluid

First find v by applying the formula for flow rate as;

Q = v*A   where ;

Q= fluid flow rate

v = flow velocity

A= cross-sectional area.

A= cross-sectional vector area of the pipe given by the formula;

A= πr² = 3.14 * 4² = 50.27 in²   where r=radius of pipe obtained from the diameter given divided by 2.

Q = fluid flow rate = 105 gpm----change to m³/s as

1 gpm = 0.00006309

105 gpm = 105 * 0.00006309 = 0.006624 m³/s

A= cross-sectional vector area = 50.27 in² -------change to m² as:

1 in² = 0.0006452 m²

50.27 in² = 50.27 * 0.0006452 = 0.03243 m²

Now calculate flow velocity as;

Q =v * A

Q/A = v

0.006624 m³/s / 0.03243 m² =v

0.2043 m/s = v

Now find the dynamic pressure q given as;

q= 1/2 * ρ*v²

q= 1/2 * 1000 * 0.2043² = 20.87 Pa

True or false Osha engineering controls are the last strategy of control an employer should use for job hazards

Answers

Answer:

false

Explanation:

Propylene (C3H6) is burned with 50 percent excess air during a combustion process. Assuming complete combustion and a total pressure of 105 kPa. Determine:

a. the air-fuel ratio
b. the temperature at which the water vapor in the products will start condensing.

Answers

Answer:

44.59°c

Explanation:

Given data :

Total pressure = 105 kpa

complete combustion

A) Determine air-fuel ratio

A-F = [tex]\frac{N_{air} }{N_{fuel} } = \frac{(Nm)_{air} }{(Nm)_{c} (Nm)_{n} }[/tex]

N = number of mole

m = molar mass

A-F = [tex]\frac{(6.75*4.76)kmole * ( 29kg/mol)}{(3kmole)* 12kg/mol + (6kmol)*(1kg/mol)}[/tex]  =  22.2 kg air/fuel

hence the ratio of Fuel-air = 1 : 22.2

B) Determine the temperature at which water vapor in the products start condensing

First we determine the partial  pressure of water vapor before using the steam table to determine the corresponding saturation temp

partial pressure of water vapor

Pv = [tex]\frac{(N_{water vapor}) }{N_{pro} } * ( P_{ro} )[/tex]

N watervapor ( number of mole of water vapor ) = 3

N pro ( total number of mole of product = 3 + 3 + 2.25 + 25.28 = 33.53 kmol

Pro = 105

hence Pv = ( 3/33.53 ) * 105 =  9.39kPa

from the steam pressure table the corresponding saturation temperature to 9.39kPa =  44.59° c

Temperature at which condensing will start = 44.59°c

An equation showing the products of propylene with their mole numbers is attached below

What prevented this weld from becoming ropey?
A lower ampera
A higher voltage
The position of the weld
The stepping motion of the weld

Answers

Answer:

If I am not mistaken I believe it is a higher voltage.

Explanation:

Hope this helps

A, higher voltage is correct

If a person runs a distance of 0.7 km in 3 min, what is his average speed in kilometres/hour ​

Answers

Answer:

14 km/hour

Explanation:

You have a motor that runs at 1.5 amps from a 12 volt power source, how many watts of power is it using?

Answers

Answer:

18 Watts

Explanation:

For this problem, we simply need to understand the relationship of power to voltage and current.  This relationship is derived from Ohm's law:

Power = Voltage * Current

Given this equation, we can say the following to find the power consumption of the motor:

Power = 12volts * 1.5amps

Power = 18 Watts

Hence, the motor is consuming 18 Watts of power.

Cheers.

A well-insulated heat exchanger has one line with 2 kg/s of air at 125 kPa and 1000 K entering, and leaving at 100 kPa and 400 K. The other line has 0.5 kg/s water entering at 200 kPa and 20 °C, and leaving at 200 kPa. Calculate the exit temperature of the water and the total rate of entropy generation?

Answers

Answer:

120°C

Explanation:

Step one:

given data

T_{wi} = 20^{\circ}C

T_{Ai}=1000K

T_{Ae}= 400kPa

P_{Wi}=200kPa

P_{Ai}=125kPa

P_{We}=200kPa

P_{Ae}=100kPa

m_A=2kg/s

m_W=0.5kg/s

We know that the energy equation is

[tex]m_Ah_{Ai}+m_Wh_W=m_Ah_{Ae}+m_Wh_{We}[/tex]

making [tex]h_{We}[/tex] the subject of formula we have

[tex]h_{We}=h_{Wi}+\frac{m_A}{mW}(h_A-h_{Ae})[/tex]

from the saturated water table B.1.1 , corresponding to  [tex]T_{wi}= 20c[/tex]

[tex]h_{Wi}=83.94kJ/kg[/tex]

from the ideal gas properties of air table B.7.1 , corresponding to T=1000K

the enthalpy is:

[tex]h_{Ai}=1046.22kJ/kg[/tex]

from the ideal gas properties of air table B.7.1 corresponding to T=400K

[tex]h_{Ae}=401.30kJ/kg[/tex]

Step two:

substituting into the equation we have

[tex]h_{We}=h_{Wi}+\frac{m_A}{mW}(h_A-h_{Ae})[/tex]

[tex]h_{We}=83.94+\frac{2}{0.5}(2046.22-401.30)\\\\h_{We}=2663.62kJ/kg[/tex]

from saturated water table B.1.2 at [tex]P_{We}=200kPa[/tex]  we can obtain the specific enthalpy:

[tex]h_g=2706.63kJ/kg[/tex]

we can see that [tex]h_g>h_{Wi}[/tex], hence there are two phases

from saturated water table B.1.2 at [tex]P_{We}=200kPa[/tex]

[tex]T_{We}=120 ^{\circ} C[/tex]

A flexible rectangular area measures 2.5 m X 5.0 m in plan. It supports an external load of 150 kN/m^2. Determine the vertical stress increase due to the load at a depth of 6.25 m below the corner of the footing.

Answers

Answer:

19.0476 kN/m^2

Explanation:

Given data:

Dimension of rectangular area = 2.5m x 5.0m

external load = 150 KN/m^2

load depth = 6.25 m

calculate the vertical stress increase due to load at depth 6.25

we will use the approximate method which is

[tex]V_{s} = \frac{qBL}{(B +Z)(L+Z)}[/tex]  ------- (1)

q = 150 kN/m^2

B = 2.5 m

L = 5 m

Z = 6.25 m

substitute the given values into the equation (1) above

hence the Vs ( vertical stress ) = 19.0476 kN/m^2

3 mA are flowing through an 18 V circuit, how much resistance (in kΩ) is in that circuit?

Answers

Answer:

6kΩ

Explanation:

If we assume that the entire circuit's current is 3 mA, then we can compute the resistance within the circuit with the application of Ohm's law:

V = IR

V/I = R

R = V / I

R = 18V / 3 mA

R = 6kΩ

Hence, the resistance of the circuit is 6kΩ.

Cheers.

You have a 12 volt power source running through a circuit that has 3kΩ of resistance, how many amps (in mA) can flow through the circuit?

Answers

Answer:

4mA

Explanation:

For this problem, we will simply apply Ohm's law:

V = IR

V/R = I

I = V / R

I = 12 volt / 3kΩ

I = 4mA

Hence, the current in the circuit is 4mA.

Cheers.

11. Technicians A and B are discussing
hybrid engines. Technician A says that idle
stop in HEVs and EREVs is a safety feature
for technicians. Technician B says that
high viscosity engine oil should be used
with most hybrid electric vehicles. Who is
correct?
O A. Technician A
B. Technician B
C. Neither Technician A nor B
D. Both Technicians A and B

Answers

Answer:

C. Neither Technician A nor B

Explanation:

Just took the test

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