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Ann [662]
3 years ago
9

A strain gage is carefully matched with three fixed-bridge resistors, each having a resistance of 120 2 The gage factor is 2.05,

and the bridge supply voltage is 3.00 V. The strain gage is installed on a structure and connected to the bridge with a pair of 15m leads with a resistance of 0.102/m.
(a) What will be the apparent strain before application of the load?

(b) If the structure is then strained to 800 ustrain, what will be the bridge output voltage?

Engineering
1 answer:
labwork [276]3 years ago
3 0

Answer:

Answer for the question:

A strain gage is carefully matched with three fixed-bridge resistors, each having a resistance of 120 2 The gage factor is 2.05, and the bridge supply voltage is 3.00 V. The strain gage is installed on a structure and connected to the bridge with a pair of 15m leads with a resistance of 0.102/m.

(a) What will be the apparent strain before application of the load?

(b) If the structure is then strained to 800 ustrain, what will be the bridge output voltage?"

is given in the attachment.

Explanation:

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Investigating how slime molds reproduce is an example of applied research.<br> True<br> False
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A thick casting with a thermal diffusivity of 5 x 10-6 m2/s is initially at a uniform temperature of 150oC. One surface of the c
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Answer:

T_o = 141.81 ^0C

Explanation:

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Thermal diffusivity \alpha = 5 \times 10 ^{-6} m^2/s

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Heat transfer coefficient h = ( we are to assume the imposed surface temperature ) = 20 W/m².K

Initial temperature = 150 ° C = (150+273) K = 423 K

Then coolant temperature with which the casting is exposed to = 20° C = (20+273)K = 293 K

Time = 40 seconds

Length = 20mm = 0.02 m

The objective is to determine the  temperature at the surface  at a depth of 20 mm after 40 seconds.

Bi = \dfrac{hL}{k}

Bi = \dfrac{20*0.02}{20}

Bi == 0.02

\tau = \dfrac{\alpha t}{L^2}

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\tau = 0.5

For a wall at 0.2 Bi

A_1 = 1.0311

\lambda _1 = 0.4328

Therefore;

\dfrac{T_o - T_{\infty}}{T_i - T_{\infty}}= A_1 e ^{-( \lambda_1^2 \ \tau)

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\dfrac{T_o - 293 }{423 - 293}= 1.0311 \times e ^{-( 0.0959 )

\dfrac{T_o - 293 }{130}= 1.0311 \times 0.9085

\dfrac{T_o - 293 }{130}= 0.937

T_o - 293= 0.937 \times 130

T_o - 293= 121.81

T_o = 121.81+ 293

T_o = 414.81 \ K

T_o = (414.81 - 273)^0C

T_o = 141.81 ^0C

4 0
4 years ago
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