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3 years ago

A non-linear analog force sensor outputs the following voltages for different forces.

Engineering

1 answer:

Serjik [45]3 years ago

4 0

Answer:

see explaination

Explanation:

Matlab code

Clc

F=[10, 15, 20, 25, 30];

V=[10, 20, 35, 55, 80];

figure(1) see attachment

subplot (2,1,1)

plot(F,V, 'o')

figure(1) see attachment

subplot (2,1,2)

plot(F,V, 'o')

Equation relating voltage to force is shown in a graph

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This acronym is a reminder of the most common types of hazards or injuries caused by electricity.

yulyashka [42]

Answer:

BE SAFE

Explanation:

it stands for Burns, Electrocution, Shock, Arch Flash/Arc Blast, Fire, Explosions

3 0

3 years ago

It is known that the connecting rod AB exerts on the crank BCa 2.5-kN force directed down andto the left along the centerline of

monitta

Answer:

M_c = 61.6 Nm

Explanation:

Given:

F_a = 2.5 KN

Find:

Determine the moment of this force about C for the two casesshown.

Solution:

- Draw horizontal and vertical vectors at point A.

- Take moments about point C as follows:

M_c = F_a*( 42 / 150 ) *88

M_c = 2.5*( 42 / 150 ) *88

M_c = 61.6 Nm

- We see that the vertical component of force at point A passes through C.

Hence, its moment about C is zero.

6 0

3 years ago

Water exiting the condenser of a power plant at 45 Centers a cooling tower with a mas flow rate of 15,000 kg/s. A stream of cool

MariettaO [177]

Answer: hello your question is incomplete below is the missing part

question :Determine the temperature of the cooled water exiting the cooling tower

answer : T = 43.477° C

Explanation:

Temp of water at exit = 45°C

mass flow rate of cooling tower = 15,000 kg/s

Temp of makeup water = 20°C

Assuming an atmospheric pressure of = 101.3 kPa

Determine temperature of the cooled water exiting the cooling tower

Water entering cooling tower at 45°C

Given that Latent heat of water at 45°C = 43.13 KJ/mol

Cp(wet air) = 1.005+ 1.884(y1)

where: y1 - Inlet mole ratio = (0.01257) / (1 - 0.01257) = 0.01273

Hence : Cp(wet air) = 29.145 + (0.01273) (33.94) = 29.577 KJ/kmol°C

First step : calculate the value of Q

Q = m*Cp*(ΔT) + W(latent heat)

Q = 321.6968 (29.577) (40-30) + 43.13 (18.26089)

Q = 95935.8547 KJ/s

Given that mass rate of water = 15000 kg/s

Hence the temperature of the cooled water can be calculated using the equation below

Q = m*Cp*∆T

Cp(water) = 4.2 KJ/Kg°C

95935.8547 = (15000)*(4.2)*(45 - T)

( 45 - T ) = 95935.8547/ 63000. ∴ T = 43.477° C

5 0

3 years ago

The aluminum rod (E1 = 68 GPa) is reinforced with the firmly bonded steel tube (E2 = 201 GPa). The diameter of the aluminum rod

Vsevolod [243]

Answer:

Explanation:

From the information given:

$E_1 = 68 \ GPa \\ \\ E_2 = 201 \ GPa \\ \\ d = 25 \ mm \ \\ \\ D = 45 \ mm \ \\ \\ L = 761 \ mm \\ \\ P = -88 kN$

The total load is distributed across both the rod and tube:

$P = P_1+P_2 --- (1)$

Since this is a composite column; the elongation of both aluminum rod & steel tube is equal.

$\delta_1=\delta_2$

$\dfrac{P_1L}{A_1E_1}= \dfrac{P_2L}{A_2E_2}$

$\dfrac{P_1 \times 0.761}{(\dfrac{\pi}{4}\times .0025^2 ) \times 68\times 10^4}= \dfrac{P_2\times 0.761}{(\dfrac{\pi}{4}\times (0.045^2-0.025^2))\times 201 \times 10^9}$

$P_1(2.27984775\times 10^{-8}) = P_2(3.44326686\times 10^{-9})$

$P_2 = \dfrac{ (2.27984775\times 10^{-8}) P_1}{(3.44326686\times 10^{-9})}$

$P_2 = 6.6212 \ P_1$

Replace $P_2$ into equation (1)

$P= P_1 + 6.6212 \ P_1\\ \\ P= 7.6212\ P_1 \\ \\ -88 = 7.6212 \ P_1 \\ \\ P_1 = \dfrac{-88}{7.6212} \\ \\ P_1 = -11.547 \ kN$

Finally, to determine the normal stress in aluminum rod:

$\sigma _1 = \dfrac{P_1}{A_1} \\ \\ \sigma _1 = \dfrac{-11.547 \times 10^3}{\dfrac{\pi}{4} \times 25^2}$

$\sigma_1 = - 23.523 \ MPa}$

Thus, the normal stress = 23.523 MPa in compression.

8 0

3 years ago

Select the correct text in the passage.

sineoko [7]

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3 years ago

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