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ss7ja [257]
4 years ago
11

An electric motor exerts a constant torque of 10 Nm on a grindstone mounted on its shaft. The moment of inertia of the grindston

e is I = 2.0 kgm2 . The system starts from rest. A. Determine the angular acceleration of the shaft when this torque is applied. B. Determine the kinetic energy of the shaft after 8 seconds of operation. C. Determine the work done by the motor in this time. D. Determine the average power delivered by the motor over this time interval
Physics
1 answer:
Dmitry_Shevchenko [17]4 years ago
6 0

Answer:

Angular acceleration = 5 rad /s ^2

Kinetic energy = 0.391 J

Work done = 0.391 J

P =6.25 W

Explanation:

The torque is given as moment of inertia × angular acceleration

angular acceleration = torque/ moment of inertia

= 10/2= 5 rad/ s^2

The kinetic energy is = 1/2 Iw^2

w = angular acceleration/time

=5/8= 0.625 rad /s

1/2 × 2× 0.625^2

=0.391 J

The work done is equal to the kinetic energy of the motor at this time

W= 0.391 J

The average power is = torque × angular speed

= 10× 0.625

P = 6.25 W

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Explanation:

By Snell's law we know at the left surface

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now we have

1 sin19 = \mu sin\theta_r

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now on the other surface we know that

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so again we have

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so we have

\theta_r = sin^{-1}\frac{0.33}{\mu}

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sin^{-1}\frac{0.33}{\mu} + sin^{-1}\frac{1}{\mu} = 49

By solving above equation we have

\mu = 1.645

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3 years ago
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Explanation:

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Using equations, determine the temperature, pressure and density of the air for a aircraft flying at 19.5 km. Is this aircraft s
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Answer:

a) - 72.5°c

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c) density =  0.063 kg/m^3

d) it is a subsonic aircraft

Explanation:

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Temperature at 19.5 km ( 19500 m )

T = -131 + ( 0.003 * altitude in meters )

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b) Determine pressure and density at 19.5 km altitude

Given :

Po (atmospheric pressure at sea level )  = 101kpa

R ( gas constant of air ) = 0.287 KJ/Kgk

T = -72.5°c ≈ 200.5 k

pressure = 3625.13 Pa

hence density = 0.063 kg/m^3

attached below is the remaining part of the solution

C) determine if the aircraft is subsonic or super sonic

Velocity ( v ) = \sqrt{CRT}  =  \sqrt{1.4*287*200.5 } = 283.8 m/s

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When electromagnetic fields interact with charged particles,
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3 years ago
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Please help
Allisa [31]

<u>We are given:</u>

Mass of the rocket = 10 kg

Weight of the Rocket = 100 N

Upward thrust applied by the rocket = 400 N

<u>Net upward force on the rocket:</u>

We are given that gravity pulls the rocket with a force of 100 N

Also, the rocket applied a force of 400N against gravity

Net upward force = Upward thrust - Force applied by gravity

Net upward force = 400 - 100

Net upward force = 300 N

<u>Upward Acceleration of the Rocket:</u>

From newton's second law:

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