Rect answer in the box. Spell all Words correctly

A _________ is interesting only if the statistics computed from transactions covered by the rule are different than those comput

qwelly [4]

Answer: Quantitative association rule.

Explanation:

The quantitative rules of association apply to the basic type of rules of association which exists as X and Y, with X and Y consisting of a collection of numerical and/or categorical attributes. Unlike general association laws, where both the left and right sides of the law should have categorical (nominal or discrete) attributes, a numerical attribute must be included in at least one attribute of the quantitative association rule (left or right).

7 0

4 years ago

A damped harmonic oscillator consists of a mass on a spring, with a damping force proportional to the speed of the block. If the

Murrr4er [49]

Answer:

TOTAL ENERGY = 0.74 j

Explanation:

Given data:

spring constant is 350 N/m

m = 0.24 kg

b = 0.41 kg/s

A = 0.075 M

$\omega = \sqrt{\frac{k}{m}}$

$= \sqrt{\frac{350}{0.24}}$

$y = e^{\frac{-b}{2m} t} A cos(\omega t)$

$=e^{\frac{-0.41}{2*0.24} t} cos (\sqrt{\frac{350}{0.24}} t) *0.075$

after one full cycle, mass will be at extreme point, hence K,E = 0

But total energy remain same

y after one full cycle is\

$y =e^{\frac{-0.41}{2*0.24} 2\pi \sqrt{\frac{0.24}{350}}} cos (2\pi*0.075)$

y = 0.06517 m

y = 6.517 cm

total energy $= \frac{1}{2} k y^2$

$= \frac{1}{2}* 350* 0.06517^2 = 0.742 J$

8 0

4 years ago

A piston–cylinder assembly contains 5 kg of air, initially at 2.0 bar, 30 C. The air undergoes a process to a state where the pr

Ainat [17]

Answer: wor done is 145. 06kJ

Heat transfer is 135.53kJ

Explanation:

No of moles of air = mass/molar mass = 5000g/28gmol^-1 = 172.65mol

P1 = 2bar =2*101300 =202600pa

T1 = 30° +273k = 303k

P2 =p1 = 202600pa

V2 =? T2 =?

Using pV = nRT

R = 8.314 PA m^3 mol^-1 k^-1

V1 = (172.65*8.314*303)/202600

V1 = 2.146m^3

For second state, 1.5pv = const = P1V1

V2 = (202600*2.146)/(1.5*202600)

V2 = 1.43m^3

Volume change = 2.146 - 1.43 =0.715m^3

Word done = pressure* volume change

W = 202600*0.716 = 145061.6J

= 145.061kJ

Using V1/T1 = V2/T2

T2 = V2T1/V1

=(1.43*303)/2.146 = 201.9k

For internal energy U

U = nCv*(T2 - T1)

*CV is the heat capacity at const. vol approximately 0.718J mol^-1 k^-1

U = 172.65*0.718*(201.9-303)

U = -12532.6J = -12.532kJ

The -ve means the system lost internal energy.

Q = U+W = total heat energy of system

Q = - 12.532+145.061 = 132.52 kJ

7 0

4 years ago

9. Technician A says to examine each bearing carefully for chips, pits, and scratches during servicing. Technician B says that d

Vanyuwa [196]

Answer:

a. is correct.

Explanation:

During servicing you should examine each bearing carefully for chips, pits, and scratches, so Technician A is correct.

Discoloration of a bearingbis not acceptable wear, so Technician B is false.

Combine the answers: Only Technician A is correct therefore answer a. is the right answer.

8 0

3 years ago

Read 2 more answers

prove that the heat transfer at the constant pressure is given by the enthalpy change during the process

mihalych1998 [28]

Answer:

at constant pressure, the heat flow for any process is equal to the change in the internal energy of the system plus the PV work done. Comparing the previous two equations shows that at constant pressure, the change in the enthalpy of a system is equal to the heat flow: ΔH=qp.

Explanation:

Give brainliest

5 0

3 years ago