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Alecsey [184]
3 years ago
13

A thin cylindrical shell and a solid cylinder have the same mass and radius. The two are released side by side and roll down, wi

thout slipping, from the top of an inclined plane that is 1.8 m above the ground. Find the final linear velocity of the thin cylindrical shell.The acceleration of gravity is 9.8 m/s 2 . Answer in units of m/s.
Physics
1 answer:
Vitek1552 [10]3 years ago
8 0

Answer:

Explanation:

Let the velocity be v

Total energy at the bottom

= rotational + linear kinetic energy

= 1/2 Iω² + 1/2 mv² ( I moment of inertia of shell  = mr² )

= 1/2 mr²ω² + 1/2 mv² ( v = ω r )

= 1/2 mv² +1/2 mv²

= mv²

mv² = mgh ( conservation of energy )

v² = gh

v = √gh

= √9.8 x 1.8

= 4.2 m /s

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Answer:

Δu=1300kJ/kg  

Explanation:

Energy at the initial state

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Is saturated vapor at initial pressure we have

p_{2}=200kPa\\x_{2}=1(stat.vapor)\\v_{2}=0.8858m^3/kg(tableA-5)

Process 2-3 is a constant volume process

p_{3}=100kPa\\v_{3}=v_{2}=0.8858m^{3}/kg\\u_{3}=1508.6kJ/kg(tableA-5)

The overall in internal energy

Δu=u₁-u₃

We replace the values in equation

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Δu=1300kJ/kg  

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Calculate the charge that flows through the cell in 1 minute. Each filament lamp has a power of 3 W and a resistance of 12 Ω
lapo4ka [179]

Answer:

24 Coulumbs

Explanation:

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Answer:

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