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Deffense [45]
2 years ago
15

In a study, the data you collect is Habits on a Always/Sometimes/Never scale.What is the level of measurement?

Physics
1 answer:
zzz [600]2 years ago
8 0

Answer:

Ordinal

Explanation:

There are four levels of measurement which include the nominal, ordinal, interval, and ratio. The data collected above is ordinal data as it qualifies the data and still indicates the ordering of the data. It gives the observer an idea of the range of data collected or its rating although mathematical calculations may not be done with it.

The other forms of data include the nominal which simply qualifies the data, the interval which qualifies the data but which the differences between the data can be obtained, and of course the data has no starting point. The ratio scale which is similar to the interval scale but which the ratios between the data obtained can be compared.

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A certain alarm clock ticks four times each second, with each tick representing half a period. The balance wheel consists of a t
Semenov [28]

Answer:

a. I=2.77x10^{-8} kg*m^2

b. K=4.37 x10^{-6} N*m

Explanation:

The inertia can be find using

a.

I = m*r^2

m = 0.95 g * \frac{1 kg}{1000g}=9.5x10^{-4} kg

r=0.54 cm * \frac{1m}{100cm} =5.4x10^{-3}m

I = 9.5x10^{-4}kg*(5.4x10^{-3}m)^2

I=2.77x10^{-8} kg*m^2

now to find the torsion constant can use knowing the period of the balance

b.

T=0.5 s

T=2\pi *\sqrt{\frac{I}{K}}

Solve to K'

K = \frac{4\pi^2* I}{T^2}=\frac{4\pi^2*2.7702 kg*m^2}{(0.5s)^2}

K=4.37 x10^{-6} N*m

3 0
2 years ago
Solve the equation x=3logy2 for y.
melisa1 [442]

X = 3 · log(Y²)

X = 3 · 2·log(Y)

X/6 = log(Y)

10^(X/6) = 10^log(Y)

Y = 10^(X/6)

6 0
2 years ago
Read 2 more answers
A parallel-plate capacitor is constructed of two horizontal 12.0-cm-diameter circular plates. A 1.0 g plastic bead, with a charg
marissa [1.9K]

Answer:

Please find the answer in the explanation

Explanation:

Given that A 1.0 g plastic bead, with a charge of -6.0 nC, is suspended between the two plates by the force of the electric field between them.

Since it is suspended, it must have been repelled by the bottom negative plate and trying to be attracted to the top plate.

We can therefore conclude that the upper plate, is positively charged

B.) The charge on the positive plate of parallel-plate capacitor is constructed of two horizontal 12.0-cm-diameter circular plates must be less than 6.0 nC

3 0
2 years ago
A thick steel sheet of area 100 in.2 is exposed to air near the ocean. After a one-year period it was found to experience a weig
vladimir1956 [14]

Answer:

Therefore the rate of corrosion 37.4 mpy and 0.952 mm/yr.

Explanation:

The corrosion rate is the rate of material remove.The formula for calculating CPR or corrosion penetration rate is

CPR=\frac{KW}{DAT}

K= constant depends on the system of units used.

W= weight =485 g

D= density =7.9 g/cm³

A = exposed specimen area =100 in² =6.452 cm²

K=534 to give CPR in mpy

K=87.6  to give CPR in mm/yr

mpy

CPR=\frac{KW}{DAT}

        =\frac{534\times( 485g)\times( 10^3mg/g)}{(7.9g/cm^3) \times (100in^2)\times (24h/day)\times (365day/yr)\times 1yr}

        =37.4mpy

mm/yr

CPR=\frac{KW}{DAT}

        =\frac{87.6\times (485g)\times (10^3 mg/g)}{(7.9g/cm^3)\times (100in^2)\times(2.54cm/in)^2\times (24h/day)\times (365day/yr)\times 1yr}

       =0.952 mm/yr

Therefore the rate of corrosion 37.4 mpy and 0.952 mm/yr.

3 0
2 years ago
Determine the specific volume of refrigerant-134a at 1 MPa and 50°C, using (a) the ideal-gas equation of state and (b) the gener
Andrej [43]

Answer:

( a ) The specific volume by ideal gas equation = 0.02632 \frac{m^{3} }{kg}

% Error =  20.75 %

(b) The value of specific volume From the generalized compressibility chart = 0.0142 \frac{m^{3} }{kg}

% Error =  - 34.85 %

Explanation:

Pressure = 1 M pa

Temperature = 50 °c = 323 K

Gas constant ( R ) for refrigerant = 81.49 \frac{J}{kg k}

(a). From ideal gas equation P V = m R T ---------- (1)

⇒ \frac{V}{m} = \frac{R T}{P}

⇒ Here \frac{V}{m} = Specific volume = v

⇒ v =  \frac{R T}{P}

Put all the values in the above formula we get

⇒ v = \frac{323}{10^{6} } ×81.49

⇒ v = 0.02632 \frac{m^{3} }{kg}

This is the specific volume by ideal gas equation.

Actual value = 0.021796 \frac{m^{3} }{kg}

Error =  0.02632 - 0.021796 =   0.004524 \frac{m^{3} }{kg}

% Error =  \frac{0.004524}{0.021796} × 100

% Error =  20.75 %

(b). From the generalized compressibility chart the value of specific volume

 \frac{V}{m} = v = 0.0142 \frac{m^{3} }{kg}

The actual value = 0.021796 \frac{m^{3} }{kg}

Error = 0.0142 - 0.021796 =  \frac{m^{3} }{kg}

% Error = \frac{- 0.0076}{0.021796} × 100

% Error =  - 34.85 %

3 0
2 years ago
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