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4 days ago

8

While spinning down from 500.0 rpm to rest, a solid uniform flywheel does 5.1 kJ of work. If the radius of the disk is 1.2 m, wh

at is its mass?

1 answer:

Keith_Richards [2.2K]4 days ago

7 0

Answer:

The mass will be calculated as 5.173 kg

Explanation:

The energy from the rotation of the uniform flywheel is given by E = 5.1 KJ = 5100 J

Angular speed $\omega =500rpm=\frac{2\times 3.14\times 500}{60}=52.33rad/sec$

Radius r = 1.2 m

The rotational kinetic energy of the flywheel is $E=\frac{1}{2}I\omega ^2$

Therefore, $5100=\frac{1}{2}\times I\times 52.33 ^2$

$I=3.724kgm^2$

The moment of inertia for a solid flywheel is $I=\frac{1}{2}mr^2$

Thus, the mass will equal 5.173 kg $3.724=\frac{1}{2}\times m\times 1.2^2$

$m=5.173kg$

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The positioning of the object along the principal axis relative to the concave mirror.

Explanation:

In a concave mirror, the characteristics of the image generated depend on where the object is situated in relation to the mirror. The distance from the mirror to the object positioned along the principal axis is key.

The nearer the object is to the mirror, the larger or more magnified the image will appear. For example, placing an object between the focal point and the concave mirror's pole results in a significantly larger image compared to an object placed outside the center of curvature of the mirror.

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How many electrons does it take to make 80 μc (microcoulombs) of charge?

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The charge for a single electron is 1.602*10^ -19 C

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28 days ago

An ideally efficient heat pump delivers 1000 J of heat to room air at 300 K. If it extracted heat from 260 K outdoor air, how mu

Yuliya22 [2438]

Answer:

Wnet, in, = 133.33J

Explanation:

Provided that

Pump heat QH = 1000J

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Cold temperature TL= 260K

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According to the first law of thermodynamics,

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COP(HP, rev) = 1/(1-0.867)

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COP(HP, rev) = 7.5

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Wnet, in = 1000/7.5

Wnet, in = 133.333J. QED

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<padditionally...><pbased on="" the="" first="" law="" rate="" at="" which="" heat="" is="" extracted="" from="" lower="" temperature="" reservoir="" calculated="" as="">

QL=QH-Wnet, in

QL=1000-133.333

QL=866.67J

</pbased></padditionally...>

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21 day ago

It's a snowy day and you're pulling a friend along a level road on a sled. You've both been taking physics, so she asks what you

Maru [2360]

Answer:

0.0984

Explanation:

The first diagram below illustrates a free body diagram that will aid in resolving this problem.

According to the diagram, the force's horizontal component can be expressed as:

$F_X = F_{cos \ \theta}$

Substituting 42° for θ and 87.0° for $F$

$F_X =87.0 \ N \ *cos \ 42 ^\circ$

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Meanwhile, the vertical component is:

$F_Y = Fsin \ \theta$

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$F_Y =87.0 \ N \ *sin \ 42 ^\circ$

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In resolving the vector, let A denote the components in mutually perpendicular directions.

The magnitudes of both components are illustrated in the second diagram provided and can be represented as A cos θ and A sin θ

The frictional force can be expressed as:

$f = \mu \ N$

Where;

$\mu$ is the coefficient of friction

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Also, the normal reaction (N) is calculated as mg - F sin θ

Substituting $F_Y \ for \ F_{sin \ \theta}$ . Normal reaction becomes:

$N = mg \ - \ F_Y$

By balancing the forces, the horizontal component of the force equals the frictional force.

The horizontal component is described as follows:

$F_X = \mu \ ( mg - \ F_Y)$

Rearranging the equation above to isolate $\mu$ leads to:

$\mu \ = \ \frac{F_X}{mg - F_Y}$

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$F_X \ = \ 64.65 \ N$

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Thus:

$\mu \ = \ \frac{64.65 N}{(73.0 kg)(9.8m/s^2) - (58.21 \ N)}$

$\mu = 0.0984$

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