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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.
Physics
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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Kingda Ka has a maximum speed of 57.2 m/s. Determine the height of the hill on this roller coaster. Explain to get full credit,
kodGreya [7K]

This question involves the concepts of the law of conservation of energy, kinetic energy, and potential energy.

The height of the hill is "166.76 m".

<h3>LAW OF CONSERVATION OF ENERGY:</h3>

According to the law of conservation of energy at the highest point of the roller coaster ride, that is, the hill, the whole (maximum) kinetic energy of the roller coaster is converted into its potential energy:

Maximum\ Kinetic\ Energy\ Lost = Maximum\ Potential\ Energy\ Gain\\\\&#10;\frac{1}{2}mv_{max}^2=mgh\\\\&#10;h=\frac{v_{max}^2}{2g}

where,

  • h = height of the hill = ?
  • v_{max} = maximum velocity = 57.2 m/s
  • g = acceleration due to gravity = 9.81 m/s²

Therefore,

h=\frac{(57.2\ m/s)^2}{2(9.81\ m/s^2)}\\\\&#10;

<u>h = 166.76 m</u>

Learn more about the law of conservation of energy here:

brainly.com/question/101125

7 0
3 years ago
What are some ways I can have a good wardrobe without breaking the bank?
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spin [16.1K]
Your answer is C) The speed of sound is higher in solids than in liquids. 
6 0
3 years ago
A rock is thrown at an angle of 60∘ to the ground. If the rock lands 25m away, what was the initial speed of the rock? (Assume a
Zinaida [17]

Answer:

v_0 = 16.82\ m/s

Explanation:

given,

angle at which rock is thrown = 60°

rock lands at distance,d = 25 m

initial speed of rock, = ?

In horizontal direction

distance = speed x time

d = v₀ cos 60° t

25 = v₀ cos 60° t............(1)

now,

in vertical direction

displacement in vertical direction is zero

using equation of motion

s = ut +\dfrac{1}{2}gt^2

0 =v_0 sin 60^0 t - 4.9 t^2

v_o sin 60^0 = 4.9 t

t = \dfrac{v_0 sin 60^0}{4.9}

putting the value of t in equation (1)

25 = v_0 cos 60^0\times \dfrac{v_0 sin 60^0}{4.9}

25 =\dfrac{v_0^2cos 60^0 sin 60^0}{4.9}v

v_0^2 = 282.90

v_0 = 16.82\ m/s

Hence, the initial speed of the rock is equal to 16.82 m/s

4 0
3 years ago
Attempt 2 You have been called to testify as an expert witness in a trial involving a head-on collision. Car A weighs 15151515 l
Natasha_Volkova [10]

Answer:

v = 28.98 ft / s

Explanation:

For this problem we must solve it in parts, let's start by looking for the speed of the two cars after the collision

In the exercise they indicate the weight of each car

          Wₐ = 1500 lb

          W_b = 1125 lb

Car B's velocity from v_b = 42.0 mph westward, car A travels east

let's find the mass of the vehicles

             W = mg

             m = W / g

             mₐ = Wₐ / g

             m_b = W_b / g

             mₐ = 1500/32 = 46.875 slug

             m_b = 125/32 = 35,156 slug

Let's reduce to the english system

             v_b = 42.0 mph (5280 foot / 1 mile) (1h / 3600s) = 61.6 ft / s

We define a system formed by the two vehicles, so that the forces during the crash have been internal and the moment is preserved

we assume the direction to the east (right) positive

initial instant. Before the crash

           p₀ = mₐ v₀ₐ - m_b v_{ob}

final instant. Right after the crash

           p_f = (mₐ + m_b) v

the moment is preserved

           p₀ = p_f

           mₐ v₀ₐ - m_b v_{ob} = (mₐ + m_b) v

           v = \frac{ m_a \ v_{oa} - m_b \ v_{ob}  }{ m_a +m_b}

we substitute the values

           v = \frac{ 46.875}{82.03} \ v_{oa} -  \frac{35.156}{82.03} \ 61.6

           v = 0.559 v₀ₐ - 26.40                  (1)

Now as the two vehicles united we can use the relationship between work and kinetic energy

the total mass is

              M = mₐ + m_b

              M = 46,875 + 35,156 = 82,031 slug

starting point. Jsto after the crash

              K₀ = ½ M v²

final point. When they stop

             K_f = 0

The work is

             W = - fr x

the negative sign is because the friction forces are always opposite to the displacement

Let's write Newton's second law

Axis y

           N-W = 0

           N = W

the friction force has the expression

            fr = μ N

we substitute

            -μ W x = Kf - Ko

             

            -μ W x = 0 - ½ (W / g) v²

            v² = 2 μ g x  

            v = \sqrt{ 2 \ 0.750 \ 32 \ 17.5}Ra (2 0.750 32 17.5  

            v = 28.98 ft / s

3 0
3 years ago
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