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2 years ago

Identify the measurement shown in figure 7 and state in centimeters

Engineering

1 answer:

Sav [38]2 years ago

6 0

Answer:

1.3cm

Explanation:

the arrow is 3 lines past the 1 so it is 1.3cm

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Does anyone know what this is

sammy [17]

Answer:

Looks like mold that got frosted over

Explanation:

4 0

1 year ago

Read 2 more answers

What is the least count of screw gauge?<br> (a) 0.01 cm<br> (b) 0.001 cm<br> (c) 0.1 cm<br> (d) 1 mm

Nonamiya [84]

Its 0.001

0.01 x100 = 1mm
0.001x100=0.1mm
0.1=10mm
1m

3 0

2 years ago

The vertical and horizontal poles at the traffic-light assembly are erected first. Determine the additional force and moment rea

konstantin123 [22]

Answer:

1. Az=258 lb

2. My=3440 ft lb

3. ∑Mz= 0

Explanation:

∑ Fx= Ax=0

∑ Fy= Ay=0

∑ Fz= Az-86-86-86

Az=258 lb

∑Mx=86 X 31 +Mx=0

Mx=2666 ft lb

∑My=86 X 40 +My=0

My=3440 ft lb

∑Mz= 0

6 0

2 years ago

A storage tank, used in a fermentation process, is to be rotationally molded from polyethylene plastic. This tank will have a co

NNADVOKAT [17]

Answer:

The volume up to cylindrical portion is approx 32355 liters.

Explanation:

The tank is shown in the attached figure below

The volume of the whole tank is is sum of the following volumes

1) Hemisphere top

Volume of hemispherical top of radius 'r' is

$V_{hem}=\frac{2}{3}\pi r^3$

2) Cylindrical Middle section

Volume of cylindrical middle portion of radius 'r' and height 'h'

$V_{cyl}=\pi r^2\cdot h$

3) Conical bottom

Volume of conical bottom of radius'r' and angle $\theta$ is

$V_{cone}=\frac{1}{3}\pi r^3\times \frac{1}{tan(\frac{\theta }{2})}$

Applying the given values we obtain the volume of the container up to cylinder is

$V=\pi 1.5^2\times 4.0+\frac{1}{3}\times \frac{\pi 1.5^{3}}{tan30}=32.355m^{3}$

Hence the capacity in liters is $V=32.355\times 1000=32355Liters$

3 0

2 years ago

1. The equilibrium number of vacancies in Ni at 1123 K is 4.7x1022m-3. The atomic weight and density of Ni at 1123 K are 58.69 g

Ahat [919]

Answer:

vacancy formation energy of Ni is 1.400 eV

Explanation:

given data

number of vacancies in Ni = 4.7 x $10^{22}$ $m^{-3}$

atomic weight = 58.69 g/mol

density = 8.8 g/cm³

solution

we get here N that is

N = $\frac{N_A \times \rho}{A}$ ...........1

N = $\frac{6.023\times 10^{23} \times 8.8 \times 10^6}{58.69}$

N = $9.030 \times 10^{28}$

and here no of vacancy will be

Nv = $N \times e^{\frac{-Qv}{kT}}$ .................2

put here value

$4.7 \times 10^{22} = 9.030 \times 10^28 \times e^{\frac{-Qv}{8.62\times 10^{-5}\times 1123}}$

$10^{-7} \times 5.20487 = e^{\frac{-Qv}{0.0968}}$

take ln both side

$ln (10^{-7} \times 5.20487 ) = ln (e^{\frac{-Qv}{0.0968}})$

$-14.468 = \frac{-Qv}{0.0968}$

Qv = 1.400 eV

so vacancy formation energy of Ni is 1.400 eV

3 0

2 years ago

Identify the measurement shown in figure 7 and state in centimeters ​

Identify the measurement shown in figure 7 and state in centimeters