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8090 [49]
3 years ago
10

Given that coal used by electric power plants has a heating value of 27.5 million btus metric ton (25 million btus per ton), det

ermine the number of kilograms (and pounds) of coal needed to produce 1 million btus of heat.
Physics
1 answer:
fredd [130]3 years ago
3 0

Answer:

• 36.4 kg of coal.

• 80 pounds of coal.

Explanation:

Using proportionality constant,

Mass of coal = 1,000,000/27,500,000 btus/metric ton

= 0.0364 metric tons of coal

Mass of coal = 1,000,000/25,000,000 btus/ton

= 0.04 tons of coal.

Converting metric tons to kilogram,

1 metric ton = 1000kg,

0.0364 metric ton;

= 36.4 kg of coal.

Converting tons to pounds,

1 ton = 2000 pounds,

0.04 metric ton;

= 80 pounds of coal.

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Answer:

Explanation:

Given that,

When Mass of block is 12kg

M = 12kg

Block falls 3m in 4.6 seconds

When the mass of block is 24kg

M = 24kg

Block falls 3m in 3.1 seconds

The radius of the wheel is 600mm

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We want to find the moment of inertia of the flywheel

Taking moment about point G.

Then,

Clockwise moment = Anticlockwise moment

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Relationship between angular acceleration and linear acceleration

a = αR

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Case 1, when y = 3 t = 4.6s

M = 12kg

Using equation of motion

y = ut + ½at², where u = 0m/s

3 = ½a × 4.6²

3 × 2 = 4.6²a

a = 6 / 4.6²

a = 0.284 m/s²

M•g•R - Mf = I•a / R + M•a•R

12 × 9.81 × 0.6 - Mf = I × 0.284/0.6 + 12 × 0.284 × 0.6

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0.4726•I + Mf = 70.632-2.0448

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Second case

Case 2, when y = 3 t = 3.1s

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Using equation of motion

y = ut + ½at², where u = 0m/s

3 = ½a × 3.1²

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a = 6 / 3.1²

a = 0.6243 m/s²

M•g•R - Mf = I•a / R + M•a•R

24 × 9.81 × 0.6 - Mf = I × 0.6243/0.6 + 24 × 0.6243 × 0.6

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Re arrange

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Solving equation 1 and 2 simultaneously

Subtract equation 1 from 2,

Then, we have

1.0406•I - 0.4726•I = 132.274 - 68.5832

0.568•I = 63.6908

I = 63.6908 / 0.568

I = 112.13 kgm²

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