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Mademuasel [1]
3 years ago
8

If it is desired to lay off a distance of 10,000' with a total error of no more than ± 0.30 ft. If a 100' tape is used and the

distance can be measured using full tape measures, what is the maxim error per tape measure allowed?
Engineering
1 answer:
Ira Lisetskai [31]3 years ago
3 0

Answer:

± 0.003 ft

Explanation:

Since our distance is 10,000 ft and we need to use a full tape measure of 100 ft. We find that 10,000 = 100 × 100.

Let L' = our distance and L = our tape measure

So, L' = 100L

Now by error determination ΔL' = 100ΔL

Now ΔL' = ± 0.30 ft

ΔL = ΔL'/100

= ± 0.30 ft/100

= ± 0.003 ft

So, the maxim error per tape is ± 0.003 ft

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8 0
3 years ago
The wall of drying oven is constructed by sandwiching insulation material of thermal conductivity k = 0.05 W/m°K between thin me
masha68 [24]

Answer:

86 mm

Explanation:

From the attached thermal circuit diagram, equation for i-nodes will be

\frac {T_ \infty, i-T_{i}}{ R^{"}_{cv, i}} + \frac {T_{o}-T_{i}}{ R^{"}_{cd}} + q_{rad} = 0 Equation 1

Similarly, the equation for outer node “o” will be

\frac {T_{ i}-T_{o}}{ R^{"}_{cd}} + \frac {T_{\infty, o} -T_{o}}{ R^{"}_{cv, o}} = 0 Equation 2

The conventive thermal resistance in i-node will be

R^{"}_{cv, i}= \frac {1}{h_{i}}= \frac {1}{30}= 0.033 m^{2}K/w Equation 3

The conventive hermal resistance per unit area is

R^{"}_{cv, o}= \frac {1}{h_{o}}= \frac {1}{10}= 0.100 m^{2}K/w Equation 4

The conductive thermal resistance per unit area is

R^{"}_{cd}= \frac {L}{K}= \frac {L}{0.05} m^{2}K/w Equation 5

Since q_{rad}  is given as 100, T_{o}  is 40 T_ \infty  is 300 T_{\infty, o}  is 25  

Substituting the values in equations 3,4 and 5 into equations 1 and 2 we obtain

\frac {300-T_{i}}{0.033} +\frac {40-T_{i}}{L/0.05} +100=0  Equation 6

\frac {T_{ i}-40}{L/0.05}+ \frac {25-40}{0.100}=0

T_{i}-40= \frac {L}{0.05}*150

T_{i}-40=3000L

T_{i}=3000L+40 Equation 7

From equation 6 we can substitute wherever there’s T_{i} with 3000L+40 as seen in equation 7 hence we obtain

\frac {300- (3000L+40)}{0.033} + \frac {40- (3000L+40)}{L/0.05}+100=0

The above can be simplified to be

\frac {260-3000L}{0.033}+ \frac {(-3000L)}{L/0.05}+100=0

\frac {260-3000L}{0.033}=50

-3000L=1.665-260

L= \frac {-258.33}{-3000}=0.086*10^{-3}m= 86mm

Therefore, insulation thickness is 86mm

8 0
3 years ago
Can I get an answer to this question please
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Answer:

  (i) 12 V in series with 18 Ω.

  (ii) 0.4 A; 1.92 W

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Explanation:

<h3>(i)</h3>

As seen by the load, the equivalent source impedance is ...

  10 Ω + (24 Ω || 12 Ω) = (10 +(24·12)/(24+12)) Ω = 18 Ω

The open-circuit voltage seen by the load is ...

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The Thevenin's equivalent source seen by the load is 12 V in series with 18 Ω.

__

<h3>(ii)</h3>

The load current is ...

  (12 V)/(18 Ω +12 Ω) = 12/30 A = 0.4 A . . . . load current

The load power is ...

  P = I^2·R = (0.4 A)^2·(12 Ω) = 1.92 W . . . . load power

__

<h3>(iii)</h3>

10 minutes is 600 seconds. At the rate of 1.92 J/s, the electrical energy delivered is ...

  (600 s)(1.92 J/s) = 1,152 J

__

<h3>(iv)</h3>

The load resistance that will draw maximum power is equal to the source resistance: 18 Ω. This is the conclusion of the Maximum Power Transfer theorem.

The power transferred to 18 Ω is ...

  ((12 V)/(18 Ω +18 Ω))^2·(18 Ω) = 144/72 W = 2 W

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Answer:

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Answer:

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