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2 years ago

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A hoop and a disk with uniform mass distribution have the same radius but the total masses are not known. Can they both roll dow

n a ramp without slipping and reach the bottom in the same time? And if so, what can you deduce about the relative masses?
a) The hoop and disk have the same mass.
b) The hoop is heavier, twice the mass of the disk.
c) The disk is heavier, twice the mass of the hoop.
d) They cannot reach the bottom at the same time regardless of mass.
e) The disk is lighter, three-fourth the mass of the hoop

1 answer:

ser-zykov [4K]2 years ago

5 0

Answer:

Explanation:

radius of hoop and the radius of disk is same = R

Let the mass of hoop is M and the mass of disk is M'.

As they reach the bottom of teh surface in same time so they travel equal distance thus, they have same acceleration.

The acceleration is given by

$a=\frac{gSin\theta }{1+\frac{I}{MR^{2}}}$

As the acceleration is same so that the moment of inertia is also same.

Moment of inertia of disk = moment of inertia of hoop

1/2 x mass of disk x R² = mass of hoop x R²

So, mass of disk = 2 x mass of hoop

Option (c) is correct.

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A frictionless piston-cylinder device contains 10 kg of superheated vapor at 550 kPa and 340oC. Steam is then cooled at constant

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a) the work (W) done during the process is -2043.25 kJ

b) the work (W) done during the process is -2418.96 kJ

Explanation:

Given the data in the question;

mass of water vapor m = 10 kg

initial pressure P₁ = 550 kPa

Initial temperature T₁ = 340 °C

steam cooled at constant pressure until 60 percent of it, by mass, condenses; x = 100% - 60% = 40% = 0.4

from superheated steam table

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so the properties of steam at p₂ = 550 kPa, and dryness fraction

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Now, work done during the process;

W = mP₁( v₂ - v₁ )

W = 10 × 550( 0.1377 - 0.5092 )

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Therefore, the work (W) done during the process is -2043.25 kJ

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b)

What would the work done be if steam were cooled at constant pressure until 80 percent of it, by mass, condenses

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v₂ = 0.001097 + 0.2( 0.34261 - 0.001097 )

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Now, work done during the process will be;

W = mP₁( v₂ - v₁ )

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Therefore, the work (W) done during the process is -2418.96 kJ

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2 years ago

A rotating flywheel has moment of inertia 18.0 kg⋅m^2 for an axis along the axle about which the wheel is rotating. Initially th

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Answer:

The rotational kinetic energy takes 0.430 seconds to become half its initial value.

Explanation:

By the Principle of Energy Conservation and the Work-Energy Theorem we know that flywheel slow down due to the action of non-conservative forces (i.e. friction), the energy losses are equal to the change in the rotational kinetic energy. That is:

$\Delta E = K_{1}-K_{2}$ (1)

Where:

$\Delta E$ - Energy losses, measured in joules.

$K_{1}$ , $K_{2}$ - Initial and final rotational kinetic energies, measured in joules.

By definition of rotational kinetic energy, we expand the equation above:

$\Delta E = \frac{1}{2}\cdot I\cdot (\omega_{1}^{2}-\omega_{2}^{2})$ (2)

Where:

$I$ - Moment of inertia of the flywheel, measured in kilograms per square meter.

$\omega_{1}$ , $\omega_{2}$ - Initial and final angular speed, measured in radians per second.

If we know that $K_{1} = 30\,J$ , $K_{2} = 15\,J$ and $I = 18\,kg\cdot m^{2}$ , then the initial angular speed is:

$K_{1} = \frac{1}{2}\cdot I \cdot \omega_{1}^{2}$ (3)

$\omega_{1}=\sqrt{\frac{2\cdot K_{1}}{I} }$

$\omega_{1} = \sqrt{\frac{2\cdot (30\,J)}{18\,kg\cdot m^{2}} }$

$\omega_{1} \approx 1.825\,\frac{rad}{s}$

$\omega_{1}\approx 0.291\,\frac{rev}{s}$

$K_{2} = \frac{1}{2}\cdot I \cdot \omega_{2}^{2}$ (4)

$\omega_{2}=\sqrt{\frac{2\cdot K_{2}}{I} }$

$\omega_{2} = \sqrt{\frac{2\cdot (15\,J)}{18\,kg\cdot m^{2}} }$

$\omega_{2} \approx 1.291\,\frac{rad}{s}$

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$t = \frac{\omega_{2}-\omega_{1}}{\alpha}$ (5)

If we know that $\omega_{1}\approx 0.291\,\frac{rev}{s}$ , $\omega_{2} \approx 0.205\,\frac{rev}{s}$ and $\alpha = -0.200\,\frac{rev}{s^{2}}$ , then the time needed is:

$t = \frac{0.205\,\frac{rev}{s}-0.291\,\frac{rev}{s}}{-0.200\,\frac{rev}{s^{2}} }$

$t = 0.43\,s$

The rotational kinetic energy takes 0.430 seconds to become half its initial value.

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