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3 years ago

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A stem and leaf display

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Crank3 years ago

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5/2055 classes displayed there’s Nooooob changes

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A coil consists of 100 turns of wire wrapped around a square frame of sides 0.25 m. The coil is centered at the origin with each

jenyasd209 [6]

Find solution in attachments below

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3 years ago

How many points do you need to define for the rectangle

Rudik [331]

Answer:

Two points

Explanation:

3 0

3 years ago

How nany degrées is the included angle of General Purpose Acme threads? A. 60 B. 29 c. 14.5 D. 10

horrorfan [7]

Answer:

B.29

Explanation:

In general purpose acme thread:

Nominal depth of thread=0.5 $ltimes Pitch$

Included angle =29 degrees

Generally Acme thread are following types

1.General purpose(G) Acme

2.Centralizing(C) Acme

3. Stub Acme

Centralizing(C) Acme threads have tighter tolerance during manufacturing as compare to General purpose(G) Acme threads.

8 0

3 years ago

An asphalt concrete mixture includes 94% aggregates by weight. The specific gravity of aggregate and asphalt are 2.7 and 1.0, re

riadik2000 [5.3K]

Answer:

The correct solution is "5.74%".

Explanation:

The given values are:

Gravity of aggregate,

$G_{agg}=2.7$

Gravity of asphalt,

$G_{asp}=1.0$

Asphalt concrete mixture,

$W_{agg}=0.94 \ W_m$

We know that,

$W_{asp}=W_m-W_{agg}$

$=0.06 \ W_m$

Now,

The theoretical specific gravity of mix,

⇒ $G_t=\frac{W_{agg}+W_{asp}}{\frac{W_{agg}}{G_{agg}} +\frac{W_{asp}}{G_{asp}} }$

By putting the values, we get

$=\frac{0.94 \ Wm+0.06 \ Wm}{\frac{0.94 \ Wm}{2.7} +\frac{0.06 \ Wm}{1} }$

$=2.45$

hence,

The percentage of voids will be:

⇒ %V = $\frac{G_t-G_m}{G_t}\times 100$

= $\frac{2.45-2.317}{2.45}\times 100$

= $\frac{0.133}{2.317}\times 100$

= $5.74$ (%)

8 0

3 years ago

Atmospheric air at 25 °C and 8 m/s flows over both surfaces of an isothermal (179C) flat plate that is 2.75m long. Determine the

vekshin1

Answer:

Re=100,000⇒Q=275.25 $\frac{W}{m^2}$

Re=500,000⇒Q=1,757.77 $\frac{W}{m^2}$

Re=1,000,000⇒Q=3060.36 $\frac{W}{m^2}$

Explanation:

Given:

For air $T_∞$ =25°C ,V=8 m/s

For surface $T_s$ =179°C

L=2.75 m ,b=3 m

We know that for flat plate

$Re$ ⇒Laminar flow

$Re>30\times10^5$ ⇒Turbulent flow

<u> Take Re=100,000:</u>

So this is case of laminar flow

$Nu=0.664Re^{\frac{1}{2}}Pr^{\frac{1}{3}}$

From standard air property table at 25°C

Pr= is 0.71 ,K=26.24 $\times 10^{-3}$

So $Nu=0.664\times 100,000^{\frac{1}{2}}\times 0.71^{\frac{1}{3}}$

Nu=187.32 ( $\dfrac{hL}{K_{air}}$ )

187.32= $\dfrac{h\times2.75}{26.24\times 10^{-3}}$

⇒h=1.78 $\frac{W}{m^2-K}$

heat transfer rate =h $(T_∞-T_s)$

=275.25 $\frac{W}{m^2}$

<u> Take Re=500,000:</u>

So this is case of turbulent flow

$Nu=0.037Re^{\frac{4}{5}}Pr^{\frac{1}{3}}$

$Nu=0.037\times 500,000^{\frac{4}{5}}\times 0.71^{\frac{1}{3}}$

Nu=1196.18 ⇒h=11.14 $\frac{W}{m^2-K}$

heat transfer rate =h $(T_∞-T_s)$

=11.14(179-25)

= 1,757.77 $\frac{W}{m^2}$

<u> Take Re=1,000,000:</u>

So this is case of turbulent flow

$Nu=0.037Re^{\frac{4}{5}}Pr^{\frac{1}{3}}$

$Nu=0.037\times 1,000,000^{\frac{4}{5}}\times 0.71^{\frac{1}{3}}$

Nu=2082.6 ⇒h=19.87 $\frac{W}{m^2-K}$

heat transfer rate =h $(T_∞-T_s)$

=19.87(179-25)

= 3060.36 $\frac{W}{m^2}$

7 0

3 years ago