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3 years ago

7

g Two Standard 1/2" B18.8.2 dowel pins are to be installed in part B with an LN1 fit. The thickness of plate A is .750 +/- .005"

The thickness of plate B is .750 +/- .005" The position tolerance of the clearance holes to one another is .014" The position tolerance of the precision holes to one another is .028" What is an appropriate MMC clearance hole diameter to allow the blocks to assemble?

1 answer:

allochka39001 [22]3 years ago

5 0

Answer:

nmuda mudaf A certain vehicle loses 3.5% of its value each year. If the vehicle has an initial value of $11,168, construct a model that represents the value of the vehicle after a certain number of years. Use your model to compute the value of the vehicle at the end of 6 years.

Explanation:

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3 years ago

1. (15) A truck scale is made of a platform and four compression force sensors, one at each corner of the platform. The sensor i

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Answer:

a). 139498.24 kg

b). 281.85 ohm

c). 10.2 ohm

Explanation:

Given :

Diameter, d = 22 m

Linear strain, $\epsilon$ = 3%

= 0.03

Young's modulus, E = 30 GPa

Gauge factor, k = 6.9

Gauge resistance, R = 340 Ω

a). Maximum truck weight

σ = Eε

σ = $0.03 \times 30 \times 10^9$

$\frac{P}{A} =0.03 \times 30 \times 10^9$

$P = 0.03 \times 30 \times 10^9\times \frac{\pi}{4}\times (0.022)^2$

= 342119.44 N

For the four sensors,

Maximum weight = 4 x P

= 4 x 342119.44

= 1368477.76 N

Therefore, weight in kg is $m=\frac{W}{g}=\frac{1368477.76}{9.81}$

m = 139498.24 kg

b). Change in resistance

$k=\frac{\Delta R/R}{\Delta L/L}$

$\Delta R = k. \epsilon R$ , since $\epsilon= \Delta L/ L$

$\Delta R = 6.9 \times 0.03 \times 340$

$\Delta R = 70.38$ Ω

For 4 resistance of the sensors,

$\Delta R = 70.38 \times 4 = 281.52$ Ω

c). $k=\frac{\Delta R/R}{\epsilon}$

If linear strain,

$\frac{\Delta R}{R} \approx \frac{\Delta L}{L}$ , where k = 1

$\Delta R = \frac{\Delta L}{L} \times R$

$\Delta R = 0.03 \times 340$

$\Delta R = 10.2$ Ω

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3 years ago

A sinusoidal voltage source produces the waveform, v t = 1 + cos 2πft. Design a system with v t as its input such that an LED wi

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Answer:

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See attached file for detailed solution of the given problem.

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3 years ago

How to find Catenary length in sag

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Answer:

buter uwu heheheheh joke

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2 years ago

Consider a very long rectangular fin attached to a flat surface such that the temperature at the end of the fin is essentially t

kodGreya [7K]

Answer:

$\frac{T-20}{130-20}= e^{-14.28*0.05}$

And if we solve for T we got:

$T= 20 + 110e^{-14.28*0.05} = 73.86 C$

The answer for this case would be $T = 73.86 C$ at 5cm from the base of the fin.

Explanation:

Data given

For this case we have the following data given:

$h = 20 \frac{W}{m^2 K}$ represent the heat transfer coefficient.

$p$ represent the perimeter for this case and would be given by:

$p = 2*0.05m +2*0.001m= 0.102m$

$k = 200 \frac{W}{m C}$ represent the thermal conductivity

$w = 5cm =0.05 m$ represent the width

$h = 1mm =0.001m$ represent the thickness

$A= wh= 0.05m *0.001m = 0.00005 m^2$

Solution to the problem

For this case we assume that we have steady conditions, the temperature of the fins varies just in one direction, the heat transfer coefficient not changes with the time and the thermal properties of the fin not change.

We can determine the temperature if the fin at x=5 cm=0.05 m from the base with the following formula:

$\frac{T-T_{\infty}}{T_b -T_{\infty}} = e^{-mx}$

Where m is a coefficient given by:

$m = \sqrt{\frac{hp}{kA}}=\sqrt{\frac{20 W/m^2 C 0.102 m}{200 W/ mC 0.00005 m^2}}= 14.28 m^{-1}$

The value of x for this case represent the distance $x =5 cm =0.05m$

$T_b =130 C$ represent the base temperature

$T_{\infty}= 20$ represent the temperature of the sorroundings or the ambient.

If we replace we have this:

$\frac{T-20}{130-20}= e^{-14.28*0.05}$

And if we solve for T we got:

$T= 20 + 110e^{-14.28*0.05} = 73.86 C$

The answer for this case would be $T = 73.86 C$ at 5cm from the base of the fin.

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3 years ago