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anastassius [24]
3 years ago
13

The second floor of a house is 6m above street level. How much work is reguired to lift a 300 kg refrigerator to the second stor

y?
Physics
1 answer:
zubka84 [21]3 years ago
8 0

Answer: 17.658kJ

Explanation:

The work done in lifting the refridgerator is given my

W = F.d

Where W = Work done, F = Force and d = distance/height to be covered by that force = 6m

But F = mg (this work is done against gravity)

Where m = mass of fridge and g = acceleration due to gravity.

F = 300kg × 9.81m/s^2

= 2943N

W = Fh (where h is the height to be scaled by the force)

W = 2943 × 6

W = 17658J

17.658kJ of work is needed to lift the refridgerator to a height of 6m

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F = m a                                  ---> 1

While in magnetic forces, force can also be expressed as:

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Lapatulllka [165]

This question is incomplete, the complete question is;

A monatomic gas fills the left end of the cylinder in the following figure. At 300 K , the gas cylinder length is 14.0 cm and the spring is compressed by65.0 cm . How much heat energy must be added to the gas to expand the cylinder length to 16.0 cm ?

Answer:

the required heat energy is 16 J

Explanation:

Given the data in the question;

Lets consider the ideal gas equation;

PV = nRT

from the image, we calculate initial pressure;

Pi = ( 2000N/M × 0.06m) / 0.0008 m²

Pi = 15 × 10⁴ Pa

next we find Initial velocity

Vi = (0.0008 m²)(0.14) = 1.1 × 10⁻⁴ m²

now we find the number of moles

n = [(15 × 10⁴ Pa)(1.1 × 10⁻⁴ m²)] / 8.31 J/molK × 300K

N = 6.6 × 10⁻³ mol

next we calculate the final temperature;

Pf = ( 2000N/m × 0.08) / 0.0008 m²

Pf = 2 × 10⁵ Pa

Calculate the final Volume

Vf = (0.0008 m² × 0.16 m = 1.28 × 10⁻⁴ m³

we also determine the final temperature

T_{f} =  (2 × 10⁵ Pa × 1.28 × 10⁻⁴ m³) / 6.6 × 10⁻³ × 8.31 J/molK

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we then calculate the change in thermal energy

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ΔU = 13.761 J

C is the isochoric molar specific heat which is equal to 3R/2 for monoatomic

now we calculate the work done;

W = 1/2 × K( x_{i\\}² - x_{f\\}² )

W = 1/2 × ( 2000 N/m) ( 0.06² - 0.08² )

= - 2.8 J

and we then calculate the heat energy using the following expression;

Q = ΔU - W

we substitute

Q = 13.761 - (- 2.8 J)

Q = 13.761 + 2.8 J)

Q =  16 J

Therefore, the required heat energy is 16 J

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