Moment of inertia for particle – problems and solutions

Moment of inertia for particle – problems and solutions

1. Two balls connected by a rod, as shown in the figure below. Ignore rod’s mass. Mass of ball P is 600 gram and mass of ball Q is 400 gram. What is the moment of inertia of the system about AB?

Known :

The axis of rotation is AB.Moment of inertia for particle – problems and solutions 1

mp = 600 gram = 0.6 kg, mq = 400 gram = 0.4 kg

rp = 20 cm = 0.2 m, rq = 50 cm = 0.5 m

Wanted : The moment of inertia of the system

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Net work Gravitational potential energy Kinetic energy – Problems and Solutions

4 Net work Gravitational potential energy Kinetic energy – Problems and Solutions

1. A 5-kg object at the height of 10-meters above the ground. Acceleration due to gravity is 10 m/s2. What is the work done on the object to moves it upward to the height of 15-meters above the ground?

Known :

Mass of object (m) = 5 kg

Height (h) = 10 meters

Acceleration due to gravity (g) = 10 m/s2

Wanted: Work was done on the object to moves it upward to the height of 15-meters above the ground.

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Moment of force – problems and solutions

Moment of force – problems and solutions

1. If FR is the net force of F1, F2, and F3, what is the magnitude of force F2 and x?

Known :

Net force (FR) = 40 NMoment of force – problems and solutions 1

Force 1 (F1) = 10 N

Force (F3) = 20 N

Wanted: The magnitude of force F2 and distance of x

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Boyle’s law Charles’s law Gay-Lussac’s law – Problems and Solutions

14 Boyle’s law Charles’s law Gay-Lussac’s law – Problems and Solutions

1. The air pressure of car’s tire is 432 kPa with the temperature of 15 oC. After several hours, the pressure of air is 492 kPa. If expansion ignored, what is the temperature of air in the tire?

Known :

Initial pressure (P1) = 432 kPa

Final pressure (P2) = 492 kPa

Initial temperature (T1) = 15 oC

Wanted: Final temperature (T2)

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Carnot cycle – problems and solutions

Carnot cycle – problems and solutions

1. If heat absorbed by the engine (Q1) = 10,000 Joule, what is the work done by the Carnot engine?

Known:Carnot cycle – problems and solutions 1

Low temperature (T2) = 400 K

High temperature (T1) = 800 K

Heat input (Q1) = 10,000 Joule

Wanted: Work done by Carnot engine (W)

Solution:

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Manometer tube – problems and solutions

Manometer tube – problems and solutions

1. A manometer tube is filled with two type of liquids. The density of liquid 1 is ρ1 = 0.8 g.cm-3, and the density of liquid 2 is ρ2 = 1 g.cm-3, and height h1 = 10 cm, then what is the height of h2.

Known :Manometer tube – problems and solutions 1

Density of liquid 1 = 0.8 g.cm-3

Density of liquid 2 = 1 g.cm-3

Height 1 = 10 cm

Wanted: Height 2

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Mechanical energy – problems and solutions

Mechanical energy – problems and solutions

The work-mechanical energy principle

1. The coefficient of the kinetic friction between block and floor (μk) is 0.5. What is the displacement of an object (s)? Acceleration due to gravity is 10 m/s2.

Known :Mechanical energy – problems and solutions 1

The coefficient of the kinetic friction (μk) = 0.5

Mass of block (m) = 4 kg

Acceleration due to gravity (g) is 10 m/s2

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Temperature and heat – problems and solutions

Temperature and heat – problems and solutions

1. On a thermometer X, the freezing point of water at -30o and the boiling point of water at 90o. 60OX = ….. oC.

Known :

The freezing point of water = -30o

The boiling point of water = 90o

Wanted : 60oX = ….. oC

Solution :

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Astronomical telescopes – problems and solutions

Astronomical telescopes – problems and solutions

1.

Astronomical telescopes – problems and solutions 1

Based on the graph above, determine the telescope magnification when the viewing eye is relaxed.

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Thermodynamics – problems and solutions

Thermodynamics – problems and solutions

The first law of thermodynamics

1. Based on graph P-V below, what is the ratio of the work done by the gas in the process I, to the work done by the gas in the process II?

Known :Thermodynamics – problems and solutions 1

Process 1 :

Pressure (P) = 20 N/m2

Initial volume (V1) = 10 liter = 10 dm3 = 10 x 10-3 m3

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