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Oduvanchick [21]
3 years ago
6

A horizontal angle was measured by repetition six times with a total station. If the initial display reading was 21o33'18" and t

he final reading was 129o20'04", determine the value of the angle to the nearest second.
Engineering
1 answer:
KIM [24]3 years ago
4 0

Answer:

The value of the angle is  107° 46' 46''

Explanation:

we know that

To find out the value of the angle , subtract the initial display reading from the final reading

Remember that

1°=60'

1'=60''

1°=3,600''

we have

Initial display reading=21° 33' 18''

Convert to seconds

21(3,600)+33(60)+18=77,598''

Final display reading=129° 20' 04''

Convert to seconds

129(3,600)+20(60)+4=465,604''

Find the difference

465,604''-77,598''=388,006''

Convert 388,006'' to degrees

388,006''/3,600=107.7794°

Convert 0.7794° to minutes

0.7794(60)=46.764'

Convert 0.764' to seconds

0.764(60)=45.84''=46''

therefore

The value of the angle is

107° 46' 46''

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provides steady-state operating data for a solar power plant that operates on a Rankine cycle with Refrigerant 134a as its worki
Vaselesa [24]

Answer:

hello some parts of your question is missing attached below is the missing part ( the required fig and table )

answer : The solar collector surface area = 7133 m^2

Explanation:

Given data :

Rate of energy input to the collectors from solar radiation = 0.3 kW/m^2

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Determine the solar collector surface area

The solar collector surface area = 7133 m^2

attached below is a detailed solution of the problem

8 0
3 years ago
A road has a crest curve, where the PVI station is a 71 35. The road transitions from a 2.1% grade to a -3.4% grade. The highest
sveticcg [70]

Answer:

Stat PVC = Stat(82+98.5)

Stat PVT = Stat(59+71.5)

Explanation

PVI = 71 + 35

Let G1 = Grade 1; G2 = Grade 2

G1 = +2.1% ; G2 = -3.4%

Highest point of curve at station = 74 + 10

General equation of a curve:

y = ax^{2} +bx+c\\dy/dx=2ax+b\\

At highest point of the curve dy/dx=o

2ax+b=0\\x=-b/2a\\x=G1L/(G2-G1)\\x=L/2 +(stat 74+10)-(stat 71+35)\\x=L/2 + 275

-G1L/(G2-G1) = (L/2 + 275)/100\\L = -2327 ft\\Station PVC = Stat(71+35)+(-2327/2)\\\\Stat PVC = 7135-1163.5\\Stat PVC = Stat(82+98.5)\\

Station PVT

Station PVT = Stat PVI + (L/2)\\Station PVT = Stat(71+35)+(-2327/2)\\Station PVT = 7135-1163.5\\Stat PVT = Stat(59+71.5)

3 0
3 years ago
An LED camping headlamp can run for 18 hours, powered by three AAA batteries. The batteries each have a capacity of 1000 mAh, an
KIM [24]

Answer:

a) the power consumption of the LEDs is 0.25 watt

b) the LEDs drew 0.0555 Amp current

Explanation:

Given the data in the question;

Three AAA Batteries;

<---- 1000mAh [ + -] 1.5 v ------1000mAh [ + -] 1.5 v --------1000mAh [ + -] 1.5 v------

so V_total = 3 × 1.5 = 4.5V

a) the power consumption of the LEDs

I_battery = 1000 mAh / 18hrs    { for 18 hrs}

I_battery = 1/18 Amp    { delivery by battery}

so consumption by led = I × V_total

we substitute

⇒ 1/18 × 4.5

P = 0.25 watt

Therefore the power consumption of the LEDs is 0.25 watt

b) How much current do the LEDs draw

I_Draw = I_battery = 1/18 Amp = 0.0555 Amp

Therefore the LEDs drew 0.0555 Amp current

5 0
3 years ago
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