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dmitriy555 [2]
3 years ago
15

List 10 uses for surveying in areas other than land surveying. Select all that apply. a. topographic surveying b. satellite surv

eying c. aerial surveying d. optical tooling e. marketing surveying f. control surveying g. as-built surveying h. statistical surveying i. telephone surveying j. construction surveying k. alignment surveying l. mine surveying m. solar surveying
Engineering
1 answer:
bazaltina [42]3 years ago
5 0

Answer:

b. satellite surveying

c. aerial surveying

d. optical tooling

e. marketing surveying

f. control surveying

h. statistical surveying

i. telephone surveying

k. alignment surveying

l. mine surveying

m. solar surveying

Explanation:

A Survey is an act of examination of the features of a subject or material under consideration. Land surveying refers to the examination of the natural and man-made features of a piece of land using scientific and mathematical methods.

Land surveying finds application in construction where a survey is made on all the structures found in a constructed property. Topographic surveying deals with examining the natural and man-made feature of a piece of land. As-built survey as the name implies examines the features and location of a building during or recently after construction. These three are examples of land surveys.

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A batch of 1000 is split into 10 smaller batches of equal size 100. The processing time of each unit is 2
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The lead time of the actual batch will be in

  • 2950 in minutes

<h3>What is Processing Time?</h3>

This refers to the amount of time which is taken for a processor to run a procedure and return a result.

We can see that a batch of 1000 is split so that they each have 10 smaller batches which has an equal size of 100 each, then if the processing time is 2 mins per machine and the set up time is 30 mins.

Hence, when this batch is processed over a serial line of 5 machines, then the lead time of the actual batch would be 2950 in minutes

Read more about processing time here:

brainly.com/question/18444145

4 0
2 years ago
Find the Rectangular form of the following phasors?
almond37 [142]

Answer:

The angles are missing in the question.

The angles are :

45,     30,    60,     90,    -34,     -56,      20,     -42,  -65,    -15

P=10, P=5,  P=25, P=54, P=65, P=95, P=250, P=8, P=35, P=150

Explanation:

1. P = 10,   θ = 45°  rectangular coordinates

x = r cosθ  ,   y = r sinθ

So, rectangular form is x + iy

x = P cosθ = 10 cos 45°

  = 7.07

y =P sinθ = 10 sin 45°

  = 7.07

Therefore, rectangular form

x + iy = 7.07 + i (7.07)

2. P = 5 , θ = 30°

x = 5 cos  30° = 4.33

y = 5 sin  30° = 2.5

So, (x+iy) = 4.33 + i (2.5)

3. P = 25 , θ = 60°

x = 25 cos  60° = 12.5

y = 25 sin  60° = 21.65

So, (x+iy) = 12.5 + i (21.65)

4. P = 54 , θ = 90°

x = 54 cos  90° = 0

y = 54 sin  90° = 54

So, (x+iy) = 0+ i (54)

5. P = 65 , θ = -34°

x = 65 cos  (-34°) = 53.88

y = 65 sin  (-34°) = -36.34

So, (x+iy) = 53.88 - i (36.34)

6. P = 95 , θ = -56°

x = 95 cos  (-56)° = 53.12

y = 95 sin  (-56)° = -78.75

So, (x+iy) = 53.12 - i (78.75)

7. P = 250 , θ = 20°

x = 250 cos  20° = 234.92

y = 250 sin 20° = 85.5

So, (x+iy) = 234.92 + i (85.5)

8. P = 8 , θ = (-42)°

x = 8 cos  (-42)° = 5.94

y = 8 sin  (-42)° = -5.353

So, (x+iy) = 5.94 - i (5.353)

9. P = 35 , θ = (-65)°

x = 35 cos  (-65)° = 14.79

y = 35 sin  (-65)° = -31.72

So, (x+iy) = 14.79 - i (31.72)

10. P = 150 , θ = (-15)°

x = 150 cos  (-15)° = 144.88

y = 150 sin  (-15)° = -38.82

So, (x+iy) = 144.88 - i (38.82)

6 0
3 years ago
A piston-cylinder apparatus has a piston of mass 2kg and diameterof
iragen [17]

Answer:

M =2.33 kg

Explanation:

given data:

mass of piston - 2kg

diameter of piston is 10 cm

height of water 30 cm

atmospheric pressure 101 kPa

water temperature = 50°C

Density of water at 50 degree celcius is 988kg/m^3

volume of cylinder is  V = A \times h

                                       = \pi r^2 \times h

                                       = \pi 0.05^2\times 0.3

mass of available in the given container is

M = V\times d

  = volume \times density

= \pi 0.05^2\times 0.3 \times 988

M =2.33 kg

6 0
4 years ago
How does heat conduction differ from convection?
Helga [31]

Explanation:

Conduction:

     Heat transfer in the conduction occurs due to movement of molecule or we can say that due to movement of electrons in the two end of same the body. Generally,  phenomenon of conduction happens in the case of solid . In conduction heat transfer takes places due to direct contact of two bodies.

Convection:

              In convection heat transfer of fluid takes place due to density difference .In simple words we can say that heat transfer occur due to motion of fluid.

7 0
3 years ago
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If gain of the critically damped system is increased, the system will behave as a) Under damped b) Over damped c) Critically dam
Ganezh [65]

Answer:

a) Under damped

Explanation:

Given that system is critically damped .And we have to find out the condition when gain is increased.

As we know that damping ratio given as follows

\zeta =\dfrac{C}{C_c}

Where C is the damping coefficient and Cc is the critical damping coefficient.

C_c=2\sqrt{mK}

So from above we can say that

\zeta =\dfrac{C}{2\sqrt{mK}}

\zeta \alpha \dfrac{1}{\sqrt K}

From above relationship we can say when gain (K) is increases then system will become under damped system.

7 0
3 years ago
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