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Montano1993 [528]

3 years ago

14

Consider the cantilever-beam Wheatstone bridge system that has four strain gages (two in compression and two in tension). Which

of the following statements is not true: (a) the change in resistance in each gage is proportional to the applied force, (b) temperature and torsional effects are automatically compensated for by the bridge, (c) the longitudinal (axial) strain in the beam is proportional to the output voltage of the bridge, (d) a downward force on the beam causes an increase in the resistance of a strain gage placed on its lower (under) side. Final Ans: (d) Compression on a lower side gage causes an increase in its resistance.

1 answer:

Ivahew [28]3 years ago

3 0

Answer: (b) temperature and torsional effects are automatically compensated for by the bridge,

Explanation:

As can be seen, strain gages 1 and 4 are on top of the beam and strain gages 2 and 3 are on the bottom of

the beam. Therefore strain gages 1 and 4 experience a tensile strain (are stretched) and strain gages 2

and 3 experience a compressive strain. If the relationship between strain and resistance is linear, then

under some load F the changes in resistance will be

R1= R1 + dR1

R4= R4 + dR4

tensile(4)and

R ¢2= R2 -dR2

R ¢3= R3 -dR3

compressive. (5)

when the four strain gages have an equal nominal

resistance (i.e., R1 = R2 = R3 = R4 = R) then the deflection method Wheatstone bridge equation reduces to the linear equation

Using known weights, a calibration curve can be established that relates the weight W to the output

voltage on a digital meter Eo,

Eo= a0 + a1W , (4)

where a0 and a1 are some constants. Once an unknown weight is known, an unknown mass or density

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3 years ago

Two cars cover the same distance in a straight line. Car a covers the distance at a constant velocity. Car b starts from rest an

Artyom0805 [142]

a) For the motion of car with uniform velocity we have , $s = ut+\frac{1}{2}at^2$ , where s is the displacement, u is the initial velocity, t is the time taken a is the acceleration.

In this case s = 520 m, t = 223 seconds, a =0 $m/s^2$

Substituting

$520 = u*223\\ \\u = 2.33 m/s$

The constant velocity of car a = 2.33 m/s

b) We have $s = ut+\frac{1}{2} at^2$

s = 520 m, t = 223 seconds, u =0 m/s

Substituting

$520 = 0*223+\frac{1}{2} *a*223^2\\ \\ a = 0.0209 m/s^2$

Now we have v = u+at, where v is the final velocity

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v = 0+0.0209*223 = 4.66 m/s

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3 years ago

A completely inelastic collision occurs between two balls of wet putty that move directly toward each other along a vertical axi

Marrrta [24]

Answer:

h = 2.64 meters

Explanation:

It is given that,

Mass of one ball, $m_1=3\ kg$

Speed of the first ball, $v_1=20\ m/s$ (upward)

Mass of the other ball, $m_2=2\ kg$

Speed of the other ball, $v_2=-12\ m/s$ (downward)

We know that in an inelastic collision, after the collision, both objects move with one common speed. Let it is given by V. Using the conservation of momentum to find it as :

$V=\dfrac{m_1v_1+m_2v_2}{m_1+m_2}$

$V=\dfrac{3\times 20+2\times (-12)}{3+2}$

V = 7.2 m/s

Let h is the height reached by the combined balls of putty rise above the collision point. Using the conservation of energy as :

$mgh=\dfrac{1}{2}mV^2$

$h=\dfrac{V^2}{2g}$

$h=\dfrac{7.2^2}{2\times 9.8}$

h = 2.64 meters

So, the height reached by the combined mass is 2.64 meters. Hence, this is the required solution.

5 0

3 years ago

A 175-kg roller coaster car starts from rest at the top of an 18.0-m hill and rolls down the hill, then up a second hill that ha

Anni [7]

Answer:

The work done by non-conservative forces on the car from the top of the first hill to the top of the second hill is 6574.75 joules.

Explanation:

By Principle of Energy Conservation and Work-Energy Theorem we present the equations that describe the situation of the roller coaster car on each top of the hill. Let consider that bottom has a height of zero meters.

From top of the first hill to the bottom

$m\cdot g \cdot h_{1} = \frac{1}{2}\cdot m\cdot v_{1}^{2} +W_{1, loss}$ (1)

From the bottom to the top of the second hill

$\frac{1}{2}\cdot m\cdot v_{1}^{2} = m\cdot g \cdot h_{2} + \frac{1}{2}\cdot m \cdot v_{2}^{2}+W_{2,loss}$ (2)

Where:

$m$ - Mass of the roller coaster car, in kilograms.

$v_{1}$ - Speed of the roller coaster car at the bottom between the two hills, in meters per second.

$g$ - Gravitational acceleration, in meters per square second.

$h_{1}$ - Height of the first top of the hill with respect to the bottom, in meters.

$W_{1, loss}$ - Work done by non-conservative forces on the car between the top of the first hill and the bottom, in joules.

$v_{2}$ - Speed of the roller coaster car at the top of the second hill, in meters per seconds.

$h_{2}$ - Height of the second top of the hill with respect to the bottom, in meters.

$W_{2, loss}$ - Work done by non-conservative forces on the car bewteen the bottom between the two hills and the top of the second hill, in joules.

By using (1) and (2), we reduce the system of equation into a sole expression:

$m\cdot g\cdot h_{1} = m\cdot g\cdot h_{2} + \frac{1}{2}\cdot m \cdot v_{2}^{2} + W_{loss}$ (3)

Where $W_{loss}$ is the work done by non-conservative forces on the car from the top of the first hill to the top of the second hill, in joules.

If we know that $m = 175\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $h_{1} = 18\,m$ , $h_{2} = 8\,m$ and $v_{2} = 11\,\frac{m}{s}$ , then the work done by non-conservative force is:

$W_{loss} = m\cdot\left[ g\cdot \left(h_{1}-h_{2}\right)-\frac{1}{2}\cdot v_{2}^{2} \right]$

$W_{loss} = 6574.75\,J$

The work done by non-conservative forces on the car from the top of the first hill to the top of the second hill is 6574.75 joules.

8 0

2 years ago