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Mumz [18]
2 years ago
9

You want to determine whether the race of the defendant has an impact on jury verdicts. You assign participants to watch a trial

with either a Hispanic or Caucasian defendant, and you measure on a 1 (not at all) to 7 (very) scale how confident they feel the defendant is guilty. This is a confidence scale represents a _________ scale of measurement
Engineering
1 answer:
Andru [333]2 years ago
6 0

Answer:

The confidence scale represents an ordinal scale of measurement

Explanation:

An ordinal scale or level of measurement is used to measure attributes that can be ranked or ordered, but the interval between the attributes do not have quantitative significance. In this case, the measurement was done on a scale of 1 - 7, with a "1" being; not all that race of defendant has an impact on jury verdicts and a "7" being "very" meaning that race indeed has impact on jury verdicts. Another example can be a survey carried out on the level of customer satisfaction on a particular product, with "1" most dissatisfied and "10 " representing most satisfied. In the first example, it is wrong to say that the difference between 1 being "not at all" and maybe 3 is the same as the difference between 5 and 7 which have different connotations, because the numbers are merely for tagging and not to quantify.

Other levels of measurement include:

1. Nominal: this is the simplest level of measurement and it is simply used to categorize the attributes. Example is taking a survey on gender in the categories of male, female and transgender.

2. Interval: the interval scale is used when the distance between two attributes have meanings but there is no true zero value associated with the scale.

3. Ratio: this combines all the other three levels of measurement and is used to categorize, used to show ranking, has meaningful distances between the attributes and the scale has a true zero point. Example is the measurement of temperature using the celcius scale thermometer, where there is a true zero point at 0°C and the distance between 5°C and 10°C is the same as the distance between 10°C and 15°C.

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Consider insulation on a circular pipe For the same thickness and type of insulation, the thermal resistance of the insulation i
leonid [27]

Answer:

b). The same for all pipes independent of the diameter

Explanation:

We know,

R_{conduction}=\frac{ln(\frac{r_{2}}{r_{1}})}{2\pi LK}

R_{convection}=\frac{1}{h(2\pi r_{2}L)}

From the above formulas we can conclude that the thermal resistance of a substance mainly depends upon heat transfer coefficient,whereas radius has negligible effects on heat transfer coefficient.

We also know,

Factors on which thermal resistance of insulation depends are :

1. Thickness of the insulation

2. Thermal conductivity of the insulating material.

Therefore from above observation we can conclude that the thermal resistance of the insulation is same for all pipes independent of diameter.

5 0
2 years ago
A cylindrical drum (2 ft. dia ,3 ft height) is filled with a fluid whose density is 40 lb/ft^3. Determine (a. the total volume o
Ksivusya [100]

Answer:

a)V=9.42\ ft^3

b)Mass in lb = 376.8 lb

Mass in slug = 11.71 slug

c)v=0.025\ ft^3/lb

d)w=1276 \ lb/ft.s^2

Explanation:

Given that

d= 2 ft

r= 1 ft

h= 3 ft

Density

\rho = 40\ lb/ft^3

a)

We know that volume V given as

V=\pi r^2 h

V=\pi \times 1^2\times 3

V=9.42\ ft^3

b)

Mass = Density x volume

mass =40\times 9.42\ lb

mass= 376.8 lb

We know that

1 lb = 0.031 slug

So 376.8 lb= 11.71 slug

Mass in lb = 376.8 lb

Mass in slug = 11.71 slug

c)

we know that specific volume(v) is the inverse of density.

v=\dfrac{1}{\rho}\ ft^3/lb

v=\dfrac{1}{40}\ ft^3/lb

v=0.025\ ft^3/lb

d)

Specific weight(w) is the product of density and the gravity(g).

w= ρ X g

w = 40 x 31.9

w=1276 \ lb/ft.s^2

8 0
2 years ago
2An oil pump is drawing 44 kW of electric power while pumping oil withrho=860kg/m3at a rate of 0.1m3/s.The inlet and outlet diam
Natasha2012 [34]

Answer:

\eta = 91.7%

Explanation:

Determine the initial velocity

v_1 = \frac{\dot v}{A_1}

    = \frac{0.1}{\pi}{4} 0.08^2

     = 19.89 m/s

final velocity

v_2 =\frac{\dot v}{A_2}

      = \frac{0.1}{\frac{\pi}{4} 0.12^2}

      =8.84 m/s

total mechanical energy is given as

E_{mech} = \dot m (P_2v_2 -P_1v_1) + \dot m \frac{v_2^2 - v_1^2}{2}

\dot v = \dot m v                       ( v =v_1 =v_2)

E_{mech} = \dot mv (P_2 -P_1) + \dot m \frac{v_2^2 - v_1^2}{2}

                = mv\Delta P + \dot m  \frac{v_2^2 -v_1^2}{2}

                 = \dot v \Delta P  + \dot v \rho \frac{v_2^2 -v_1^2}{2}

              = 0.1\times 500 + 0.1\times 860\frac{8.84^2 -19.89^2}{2}\times \frac{1}{1000}

E_{mech} = 36.34 W

Shaft power

W = \eta_[motar} W_{elec}

    =0.9\times 44 =39.6

mechanical efficiency

\eta{pump} =\frac{ E_{mech}}{W}

=\frac{36.34}{39.6} = 0.917  = 91.7%

8 0
2 years ago
. A belt drive is desired to couple the motor with a mixer for processing corn syrup. The 25-hp electric motor is rated at 950 r
forsale [732]

Answer:

Hello the table which is part of the question is missing and below are the table values

For a 5V belt the available diameters are : 5.5, 5.8, 5.9, 6.2, 6.3, 6.6, 12.5, 13.9, 15.5, 16.1, 18.5, 20.1

Answers:

belt size = 140 in with diameter of 20.1n

actual speed of belt = 288.49 in/s

actual center distance = 49.345 in

Explanation:

Given data :

Electric motor (driver sheave) speed (w1) = 950 rpm

Driven sheave speed (w2) = 250 rpm

pick D1 ( diameter of driver sheave)  = 5.8 in  ( from table )

To select an appropriate belt size we apply the equation for the velocity ratio to get the diameter first

VR = \frac{w1}{w2} = 950 / 250

also since the speed of  belt would be constant then ;

Vb = w1r1 = w2r2 ------- equation 1

r = d/2

substituting the value of r into equation 1

equation 2 becomes : \frac{w1}{w2} = \frac{d2}{d1}    = VR

Appropriate belt size ( d2) can be calculated as

d2 = \frac{w1d1}{w2} = \frac{950 * 5.8}{250} = 22.04

From the given table the appropriate belt size would be : 20.1 because it is the closest to the calculated value

next we have to determine the belt length /size

L = 2C + \frac{\pi }{2} ( d1+d2) + \frac{(d2-d1)^2}{4C}

inputting  all the values into the above equation including the value of C as calculated below

L ≈ 140 in

Calculating the center distance

we use this equation to get the ideal center distance

d2< C_{ideal} < 3( d1 +d2)

22.04 < c < 3 ( 5.8 + 20.1 )

22.04 < c < 77.7

the center distance is between 22.04 and 77.7  but taking an average value

ideal center distance would be ≈ 48 in

To calculate the actual center distance we use

C = \frac{B+\sqrt{B^2 - 32(d2-d1)^2} }{16} -------- equation 3

B = 4L -2\pi (d2 + d1 )

inputting all the values into (B)

B = 140(4) - 2\pi( 20.01 + 5.8 )

B ≈ 399.15 in

inputting all the values gotten Back to equation 3 to get the actual center distance

C = 49.345 in ( actual center distance )

Calculating the actual belt speed

w1 = 950 rpm = 99.48 rad/s

belt speed ( Vb) = w1r1 = w1 * \frac{d1}{2}

                           = 99.48 * 5.8 / 2 = 288.49 in/s

3 0
2 years ago
A heat pump operates on a Carnot heat pump cycle with a COP of 12.5. It keeps a space at 24°C by consuming 2.15 kW of power. Det
Vinil7 [7]

Answer:

a) T_{L} = 273.378\,K\,(0.228\,^{\textdegree}C), b) \dot Q_{H} = 26.875\,kW

Explanation:

a) The Coefficient of Performance of the Carnot Heat Pump is:

COP_{HP} = \frac{T_{H}}{T_{H}-T_{L}}

After some algebraic handling, the temperature of the cold reservoir is determined:

T_{H}-T_{L} = \frac{T_{H}}{COP_{HP}}

T_{L} = T_{H}\cdot \left(1-\frac{1}{COP_{HP}}  \right)

T_{L} = (297.15\,K)\cdot \left(1-\frac{1}{12.5}\right)

T_{L} = 273.378\,K\,(0.228\,^{\textdegree}C)

b) The heating load provided by the heat pump is:

\dot Q_{H} = COP_{HP}\cdot \dot W

\dot Q_{H} = (12.5)\cdot (2.15\,kW)

\dot Q_{H} = 26.875\,kW

4 0
2 years ago
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