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Rufina [12.5K]
3 years ago
13

The section of the area to be examined is shown circumscribed by broken lines with circles at

Engineering
1 answer:
Aloiza [94]3 years ago
3 0

This question is about Circle Geometry. it evaluates connected and broken lines with respect to circles.

<h3>What is Circle Geometry?</h3>

This refers to the body of knowledge in mathematics that has to do with the various problems associated with the Circle.

In real-world scenarios, circle geometry is used in technologies involving:

  • Camera lenses
  • Circular Architectural structures
  • Steering Wheels
  • Buttons etc.

Learn more about Circle Geometry at:
brainly.com/question/24375372

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Air flows from a large reservoir in which the pressure and temperature are 1 MPa and 30°C, respectively, through a convergent–di
SSSSS [86.1K]

Answer:

The solution is attached in the attachment.

3 0
3 years ago
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Triss [41]

Answer:

2.5 is the required details

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3 years ago
What part of the scope pattern show the duration of the spark?
dezoksy [38]

Answer: This spark energy trigger ignition and combustion in the compressed air-fuel mixture. This discharge is of extremely brief duration (about 1/1000 of a second) and is extraordinarily complex!

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2 years ago
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A long, horizontal, pressurized hot water pipe of 15cm diameter passes through a room where the air temperature is 24degree C. T
solmaris [256]

Answer:

Rate of heat transfer to the room air per meter of pipe length equals 521.99 W/m

Explanation:

Since it is given that the radiation losses from the pipe are negligible thus the only mode of heat transfer will be by convection.

We know that heat transfer by convection is given by

\dot{Q}=hA(T-T_{\infty })

where,

h = heat transfer coefficient = 10.45 W/m^{2}K (free convection in air)

A = Surface Area of the pipe

Applying the given values in the above formula we get

\dot{Q}=10.45\times \pi DL\times (130+273-(24+273))\\\\\frac{\dot{Q}}{L}=10.45\times 0.15\times \pi \times (130-24)\\\\\frac{\dot{Q}}{L}=521.99W/m

5 0
3 years ago
A hypothetical metal has an orthorhombic unit cell for which the a, b, and c lattice parameters are 0.413 nm, 0.665 nm, and 0.87
Inessa [10]

Answer:

atomic radius  R = 0.157 nm

metal atomic weight = 72.27 g/mol

Explanation:

given data

parameters a =  0.413 nm

parameters b = 0.665 nm

parameters c =  0.876 nm

atomic packing factor = 0.536

density = 3.99 g/cm³

to find out

atomic radius and  atomic weight

solution

we apply  here atomic packing factor (x) that is

atomic packing factor (x) = \frac{volume(sphere)}{volume(unit\ cell)}  ..................1

put here value we get

atomic packing factor = \frac{8*(4/3)*\pi R^3}{3*a*b*c}

R = (\frac{3(x)(abc)}{32\pi })^{1/3}

R =  (\frac{3(0.536)(0.413*0.665*0.876)}{32\pi })^{1/3}

atomic radius  R = 0.157 nm

and

now we get here metal atomic weight that is

metal atomic weight = \frac{\rho (abc)(N_A)}{no\ of\ atom}   ....................2

metal atomic weight = \frac{3.99 (0.413*0.665*0.876)(6.023*10^{23})}{8}  

metal atomic weight = 72.27 g/mol

6 0
3 years ago
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