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2 years ago

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A ________ is only achieved through control and involves a specific change in one event (dependent variable) that can reliably b

e produced by specific manipulations of another event (independent variable), and the change in the dependent variable is unlikely to be the result of other extraneous factors (confounding variables).

1 answer:

prohojiy [21]2 years ago

6 0

Answer:

Functional Relationship

Explanation:

A functional relationship is only achieved through control and involves a specific change in one event (dependent variable) that can reliably be produced by specific manipulations of another event (independent variable), and the change in the dependent variable is unlikely to be the result of other extraneous factors (confounding variables

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Water flows at a rate of 10 gallons per minute in a new horizontal 0.75?in. diameter galvanized iron pipe. Determine the pressur

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Answer:

$\frac{\delta p }{l} = 30.4 lb/ft^3$

Explanation:

Given data:

flow rate = 10 gallon per minute = 0.0223 ft^3/sec

diameter = 0.75 inch

we know discharge is given as

Q = VA

solve for velocity V = \frac{Q}{A}[/tex]

$V = \frac{0.223}{\frac{\pi}{4} \frac{0.75}{12}}$

V = 7.27 ft/sec

we know that Reynold number

$Re = \frac{VD}{\nu}$

$Re = \frac{7.27 \times \frac{0.75}{12}}{1.21\times 10^{-5}}$

$Re = 3.76 \times 10^4$

calculate the $\frac{\epsilon }{D}$ ratio to determine the fanning friction f

$\frac{\epsilon }{D} = \frac{0.0005}{\frac{0.75}{12}} = 0.008$

from moody diagram f value corresonding to Re and $\frac{\epsilon }{D}$ is 0.037

for horizontal pipe

$\delta p = \frac{f l \rho v^2}{2D}$

$\frac{\delta p }{l} = \frac{1 \times 0.037 \times 1.94 \times 7.27}{\frac{0.75}{12}}$

where 1.94 slug/ft^3is density of water

$\frac{\delta p }{l} = 30.4 lb/ft^3$

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2 years ago

a 250 pound person would use a type 1 ladder even if they were carrying a load with them true or faults

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Answer:True

Explanation:

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A milling operation was used to remove a portion of a solid bar of square cross section. Forces of magnitude P = 18 kN are appli

monitta

The smallest allowable depth is $d=16.04 \mathrm{mm}$ for the milled portion of bar.

<u>Explanation:</u>

Given,

Magnitude of force, $\mathbf{p}=18 \mathrm{kN}$

$a=30 \mathrm{mm}$

$=0.03 \mathrm{m}$

Allowable stress, $\sigma_{a l l}=135 \mathrm{MPa}$

cross sectional area of bar,

$A=a \times d$

$A=a d$

e - eccentricity

$e=\frac{a}{2}-\frac{d}{2}$

The internal forces in the cross section are equivalent to a centric force P and a bending couple M.

$M=P e$

$=P\left(\frac{a}{2}-\frac{d}{2}\right)$

$=\frac{P(a-d)}{2}$

Allowable stress

$\sigma=\frac{P}{A}+\frac{M c}{I}$

$c=\frac{d}{2}$

Moment of Inertia,

$I=\frac{b d^{3}}{12}$

$=\frac{a d^{3}}{12}$

$\therefore \sigma=\frac{P}{a d}+\frac{\frac{P(a-d)}{2} \times \frac{d}{2}}{\frac{a d^{3}}{12}}$

$\sigma=\frac{P}{a d}+\frac{3 P(a-d)}{a d^{2}}\\$

$\sqrt{x} \sigma\left(a d^{2}\right)=P d+3 P(a-d)$

$\sigma\left(a d^{2}\right)=P d+3 P a-3 P d$

$\sigma\left(a d^{2}\right)=(P-3 P) d+3 P a$

$\left(\sigma a d^{2}\right)=-2 P d+3 P a$

$\sigma d^{2}=-\frac{2 P}{a} d+3 P$

By substituting values we get,

$\left(135 \times 10^{6}\right) d^{2}+\frac{2 \times 18 \times 10^{3}}{0.03} d-3\left(18 \times 10^{3}\right)=0$

$\left(135 \times 10^{6}\right) d^{2}+\left(12 \times 10^{5}\right) d-54 \times 10^{3}=0$

On solving above equation we get, $d=0.01604 \mathrm{m}\\$

$d=16.04 \mathrm{mm}$

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2 years ago

Rigid bar ACB is supported by an elastic cir-cular strut DC having an outer diameter of 15 in. and inner diameter of 14.4 in. Th

Ne4ueva [31]

Answer:

The change in length of the circular strut DC = 0.0028 in.

The vertical displacement of the rigid bar at point B = 0.00378 in.

Explanation:

We have the following parameters or information in the question given above:

=> The outer diameter = 15 in., the inner diameter = 14.4 in., the modulus elasticity of E = 29,000 ksi, and the Point load P = 5kips.

The diagram showing the rigid bar ACB is supported by an elastic cir-cular strut DC is given in the attached picture below.

According to Newton's law of motion, it can be seen that the force on CD, that is FCD is equal and opposite to ACD. Hence, FCD = ACD.

Where FCD = p × [4 + 5] ÷ [sin Ф × 4].

kindly note that from the diagram sin Ф = 3/5, cos Ф = 4/5 and tan Ф = 3/4. Also p =5.

Hence, FCD =[ 5 × 9] ÷ [3/5 × 4] = 18.75 kip. So FCD = ACD.

The next thing here is to determine the area and length of CD, say the area of CD is G, thus, G = π/4 × [ 15² - 14.4²] = 13.854 in².

The lenght of CD is = √[4² + 3²] = √[16 + 9] = 5ft. Thus, 5 × 12 = 60in.

Hence, the change in length of the circular strut DC = [18.75 × 60] ÷ 13.854 × 29000 = 0.0028 in.

The vertical deflection of CD = 0.0028 × 3/5 = 0.00168 in.

We have that; 4 /CV = 9BV. Hence, BV = 9/4× CV.

(CV = vertical deflection of CD).

The vertical displacement of the rigid bar at point B = 9/ 4 × 0.00168 in = 0.00378 in.

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