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3 years ago

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A disk is rotating around an axis located at its center. The angular velocity is 0.5 rad/s. The radius of the disk is 0.4 m. Wha

t is the magnitude of the velocity at a point located on the outer edge of the disk, in units of m/s?

2 answers:

dimaraw [331]3 years ago

4 0

Answer:

0.2 m/s

Explanation:

The velocity of a point on the edge of a disk rotating disk can be calculated as:

$v=\omega*r$

Where $\omega$ is the angular velocity and r the radius of the disk. This leads to:

$v=0.5\,rad/s\,*\,0.4\,m=0.2\,m/s$

Anna007 [38]3 years ago

4 0

Answer:

0.2 m/s

Explanation:

<u>Step 1: </u>identify the given parameters

angular velocity, ω = 0.5 rad/s

radius of the disk, r = 0.4m

<u>Note</u>: the inner part of the disk and outer edge spin at the same rate. This means that the velocity at inner part of the disk is the same as the outer part.

<u>Step 2:</u> calculate the velocity of the disk at the outer edge in m/s

Velocity = Angular velocity (rad/s) X radius (m)

Velocity = 0.5 rad/s X 0.4 m

Velocity = 0.2 m/s

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3 years ago

A waste stabilization pond is used to treat a dilute municipal wastewater before the liquid is discharged into a river. The infl

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(b) calculate the resultant supply power factor

First convert the hp into kW

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Find the electrical power (real power) of the motor

$P_{elec} =P_{mech}/n$

where $n$ is the efficiency of the motor

$P_{elec} =186.5/0.80=233.125$ $kW$

The current in the motor is

$I_{m} =(P/\*V*pf)$

The pf of motor is 0.85 Leading

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3 years ago

An air-standard Carnot cycle is executed in a closed system between the temperature limits of 350 and 1200 K. The pressures befo

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This question is incomplete, the complete question is;

An air-standard Carnot cycle is executed in a closed system between the temperature limits of 350 and 1200 K. The pressures before and after the isothermal compression are 150 and 300 kPa, respectively. If the net work output per cycle is 0.5 kJ, determine (a) the maximum pressure in the cycle, (b) the heat transfer to air, and (c) the mass of air. Assume variable specific heats for air.

Answer:

a) the maximum pressure in the cycle is 30.01 Mpa

b) the heat transfer to air is 0.7058 KJ

c) mass of Air is 0.002957 kg

Explanation:

Given the data in the question;

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Pr1 = 238

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we substitute

Qin = 0.5 / 0.7084

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S4 - S3 = (S°4 - S°3) - R.In(P4/P3)

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W"_netout = (S2 - S1)(TH - TL)

we substitute

W"_netout = ( 0.1989 Kj/Kg.K )( 1200 - 350)K

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m = 0.002957 kg

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5 0

3 years ago