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2 years ago

7

how are the construction steps for bisecting an angle similar to the steps you would have taken to construct and bisect a line s

egment? how are they different?

1 answer:

GREYUIT [131]2 years ago

5 0

Answer:

does the question entail anything else?

Explanation:

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You installed a new 40 gallon water heater with a 54,000 BTUh burner. The underground water temperature coming into the house is

allochka39001 [22]

Answer:

For most uses you'll want your water heated to 120 F(49 C) In this example you'd need a demand water heater that produces a temperature rise and it will take about 2 hours

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3 years ago

Which cod is the best whoever has the best awnser gets brainliest

Gnesinka [82]

Explanation:

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2 years ago

): drivers must slow down from 60 to 40 mi/hr to negotiate a severe curve. A warning sign is visible for a distance of 120 ft. H

Yakvenalex [24]

Answer:

Explanation:

Total Stopping distance is the sum of the reaction distance and the braking distance, such that ;

d=dr + db =1.47St +[ ( Si 2- Sf2 ) / 30(F/0.01G) ]

In this case, the reaction time, t, is the AASHTO standard, or 2. 5 s. The friction factor, F, is based upon the standard AASHTO deceleration rate of 11.2 ft/s2 (F = 11.2/32.2 = 0.348) and the speed is given as 40 mph.

d = 1.47x60x2.5+ [ (60^2-40^2) / 30x 0.348 ]

d = 220.5 + ( (3600-1600) / 10.44 )

d = 220.5 + 191.57

d = 412.07ft.

Since the sign can be seen clearly at 120 ft.

Then the position of the sign should be,

= 412.07 - 120

= 292.07 ft

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3 years ago

Now, suppose that you have a balanced stereo signal in which the left and right channels have the same voltage amplitude, 500 mV

eimsori [14]

Answer:

R₁ = 32kΩ

Explanation:

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3 years ago

Air at 38°C and 97% relative humidity is to be cooled to 14°C and fed into a plant area at a rate of 510m3/min. (a) Calculate th

Katarina [22]

To develop the problem it is necessary to apply the concepts related to the ideal gas law, mass flow rate and total enthalpy.

The gas ideal law is given as,

$PV=mRT$

Where,

P = Pressure

V = Volume

m = mass

R = Gas Constant

T = Temperature

Our data are given by

$T_1 = 38\°C$

$T_2 = 14\°C$

$\eta = 97\%$

$\dot{v} = 510m^3/kg$

Note that the pressure to 38°C is 0.06626 bar

PART A) Using the ideal gas equation to calculate the mass flow,

$PV = mRT$

$\dot{m} = \frac{PV}{RT}$

$\dot{m} = \frac{0.6626*10^{5}*510}{287*311}$

$\dot{m} = 37.85kg/min$

Therfore the mass flow rate at which water condenses, then

$\eta = \frac{\dot{m_v}}{\dot{m}}$

Re-arrange to find $\dot{m_v}$

$\dot{m_v} = \eta*\dot{m}$

$\dot{m_v} = 0.97*37.85$

$\dot{m_v} = 36.72 kg/min$

PART B) Enthalpy is given by definition as,

$H= H_a +H_v$

Where,

$H_a$ = Enthalpy of dry air

$H_v$ = Enthalpy of water vapor

Replacing with our values we have that

$H=m*0.0291(38-25)+2500m_v$

$H = 37.85*0.0291(38-25)-2500*36.72$

$H = 91814.318kJ/min$

In the conversion system 1 ton is equal to 210kJ / min

$H = 91814.318kJ/min(\frac{1ton}{210kJ/min})$

$H = 437.2tons$

The cooling requeriment in tons of cooling is 437.2.

3 0

3 years ago