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Balakumar Engineering Diploma Tuition trainer in Chennai

Balakumar

Automobile Engineer

Tambaram West, Chennai, India - 600045.

1 Student

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Overview

I completed my M.E (Internal Combustion Engineering) in the College of Engineering, Guindy, Anna University. I had completed my B.E (Automobile Engineering) in shanmuganathan Engg College. I can teach subjects Related to Automobile Engineering & thermal sciences, the strength of Materials.

Languages Spoken

English

Tamil

Address

Tambaram West, Chennai, India - 600045

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Teaches

Engineering Diploma Tuition

Class Location

Online Classes (Video Call via UrbanPro LIVE)

Student's Home

Tutor's Home

Years of Experience in Engineering Diploma Tuition

5

Engineering Diploma Branch

Engineering Diploma 1st Year, Automobile Engineering Diploma, Mechanical Engineering Diploma

Engineering Diploma Subject

Basic Math, Engineering Mechanics, Basic Physics, Engineering Graphics, Engineering Mathematics

Mechanical Engineering Diploma Subject

Applied Mathematics, Design of Machine Elements, Thermal Engineering, Mechanical Engineering Materials, Fundamentals of Electronics, Mechanical Engineering Drawing, Strength of Materials, Power Plant Engineering, Automobile Engineering

Automobile Engineering Diploma Subject

Automobile Manufacturing Processes, Automobile Air Conditioning, Automobile component Design, Automotive Electrical and Electronic Syaytems, Applied Math, Strength of Materials, Mechanical Engineering Drawing, Vehicle aerodynamics and Design, Materials and Manufacturing Process, Heat Power Engineering, Advanced Automobile Engines, Vehicle Maintainence, Vehicle Testing, Automobile Systems, Automibile Engines, Automobile Transmission Systems

Type of class

Regular Classes

Class strength catered to

One on one/ Private Tutions

Taught in School or College

Yes

BTech Tuition
1 Student

Class Location

Online Classes (Video Call via UrbanPro LIVE)

Student's Home

Tutor's Home

Years of Experience in BTech Tuition

4

BTech Mechanical subjects

Internal Combustion Engines and Emissions, Turbomachines, Automobile Engineering, Fluid Mechanics, Engineering Drawing & Graphics, Automotive Fuels and Fuel Sysytems, Refrigeration & Air Conditioning Systems, Strength of Materials, Finite Element Analysis, Thermodynamics, Heat & Mass Transfer

BTech Branch

BTech Automobile Engineering, BTech 1st Year Engineering, BTech Mechantronics Engineering, BTech Mechanical Engineering

BTech Automobile subjects

New Generation And Hybrid Vehicles, Engineering Mechanics, Vehicle Dynamics, Autotronics, Automotive Pollution Control, Thermodynamics, Materials Science And Metallurgy, Strength Of Materials, Computational Fluid Dynamics, Automotive Engines, Engine Auxiliary Systems, Composite Materials And Mechanics, Automotive Chassis And Suspension, Engine Tribology, Two And Three Wheelers, Theory Of Machines, Ergonomics In Automotive Design, Alternative Fuels And Energy System, Automotive Electrical And Electronics System, Vehicle Aerodynamics, Automotive Safety

Type of class

Regular Classes

Class strength catered to

One on one/ Private Tutions

BTech Mechantronics subjects

Strength of Materials, Autotronics, Hybrid and Electric Vehicles, Automobile Engineering, Mechanical Vibrations, Design of Machine Elements, Engineering Thermodynamics and Heat Transfer

Taught in School or College

Yes

BTech 1st Year subjects

Engineering Physics, Basic Electronics, Engineering Graphics, Mechanics Of Solids, Basic Mechanical Engineering

Class 9 Tuition

Class Location

Online Classes (Video Call via UrbanPro LIVE)

Student's Home

Tutor's Home

Years of Experience in Class 9 Tuition

4

Board

CBSE, State

Subjects taught

Mathematics, Science, Tamil

Taught in School or College

Yes

Class 10 Tuition

Class Location

Online Classes (Video Call via UrbanPro LIVE)

Student's Home

Tutor's Home

Years of Experience in Class 10 Tuition

4

Board

CBSE, State

Subjects taught

Mathematics, Tamil, Science

Taught in School or College

Yes

Class 11 Tuition

Class Location

Online Classes (Video Call via UrbanPro LIVE)

Student's Home

Tutor's Home

Years of Experience in Class 11 Tuition

4

Board

State, CBSE

Subjects taught

Mathematics, English, Physics, Home Science, Tamil, Electronics

Taught in School or College

Yes

Teaching Experience in detail in Class 11 Tuition

4yrs

Class 12 Tuition

Class Location

Online Classes (Video Call via UrbanPro LIVE)

Student's Home

Tutor's Home

Years of Experience in Class 12 Tuition

4

Board

State, CBSE

Subjects taught

Tamil, Physics, Mathematics, Home Science, English, Electronics

Taught in School or College

Yes

Teaching Experience in detail in Class 12 Tuition

4yrs

Reviews

No Reviews yet!

FAQs

1. Which Engineering Diploma branches do you tutor for?

Engineering Diploma 1st Year, Automobile Engineering Diploma and Mechanical Engineering Diploma

2. Do you have any prior teaching experience?

Yes

3. Which classes do you teach?

I teach BTech Tuition, Class 10 Tuition, Class 11 Tuition, Class 12 Tuition, Class 9 Tuition and Engineering Diploma Tuition Classes.

4. Do you provide a demo class?

Yes, I provide a free demo class.

5. How many years of experience do you have?

I have been teaching for 5 years.

Answers by Balakumar (5)

Answered on 20/09/2018 Learn CBSE/Class 11/Science/Physics/Unit 7-Properties of Bulk Matter/Chapter 9-Mechanical Properties of Solids

Water pressure at the bottom, p = 1.1 × 108 PaInitial volume of the steel ball, V = 0.32 m3Bulk modulus of steel, B = 1.6 × 1011 Nm–2The ball falls at the bottom of the Pacific Ocean, which is 11 km beneath the surface.Let the change in the volume of the ball on reaching the bottom... ...more

Water pressure at the bottom, = 1.1 × 108 Pa
Initial volume of the steel ball, V = 0.32 m3
Bulk modulus of steel, B = 1.6 × 1011 Nm–2
The ball falls at the bottom of the Pacific Ocean, which is 11 km beneath the surface.
Let the change in the volume of the ball on reaching the bottom of the trench be ΔV.
Bulk modulus, B = p / (?V/V)
?V  =  pV/B
= 1.1 × 108 × 0.32 / (1.6 × 1011 )  =  2.2 × 10-4 m3
Therefore, the change in volume of the ball on reaching the bottom of the trench is 2.2 × 10–4 m3.

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Answered on 20/09/2018 Learn CBSE/Class 11/Science/Physics/Unit 7-Properties of Bulk Matter/Chapter 9-Mechanical Properties of Solids

Cross-sectional area of wire A, a1 = 1.0 mm2 = 1.0 × 10–6 m2Cross-sectional area of wire B, a2 = 2.0 mm2 = 2.0 × 10–6 m2Young’s modulus for steel, Y1 = 2 × 1011 Nm–2Young’s modulus for aluminium, Y2 = 7.0 ×1010Nm–2(a) Let a small mass m be suspended... ...more

Cross-sectional area of wire Aa1 = 1.0 mm= 1.0 × 10–6 m2
Cross-sectional area of wire Ba2 = 2.0 mm= 2.0 × 10–6 m2
Young’s modulus for steel, Y1 = 2 × 1011 Nm–2
Young’s modulus for aluminium, Y2 = 7.0 ×1010Nm–2

(a) Let a small mass m be suspended to the rod at a distance y from the end where wire A is attached.
Stress in the wire = Force / Area  =  F / a
If the two wires have equal stresses, then:
F1 / a1  =  F2 / a2
Where,
F1 = Force exerted on the steel wire
F2 = Force exerted on the aluminum wire
F1 / F2 = a1 / a2  =  1 / 2    ….(i)
The situation is shown in the following figure:

 
 

Taking torque about the point of suspension, we have:
F1y = F2 (1.05 – y)
F1 / F2 = (1.05 – y) / y    ……(ii)
Using equations (i) and (ii), we can write:
(1.05 – y) / y  = 1 / 2
2(1.05 – y)  =  y
y = 0.7 m
In order to produce an equal stress in the two wires, the mass should be suspended at a distance of 0.7 m from the end where wire A is attached.

(b) Young’s modulus = Stress / Strain
Strain = Stress / Young’s modulus  =  (F/a) / Y
If the strain in the two wires is equal, then:
(F1/a1) / Y1  =  (F2/a2) / Y2
F1 / F2 = a1Y1 / a2Y2
a1 / a2 = 1/2
F1 / F2 = (1 / 2) (2 × 1011 / 7 × 1010)  =  10 / 7      …….(iii)
Taking torque about the point where mass m, is suspended at a distance y1 from the side where wire A attached, we get:
F1yF2 (1.05 – y1)
F1 / F2  =  (1.05 – y1) / y1    ….(iii)
Using equations (iii) and (iv), we get:
(1.05 – y1) / y1  =  10 / 7
7(1.05 – y1)  =  10y1
y1 = 0.432 m
In order to produce an equal strain in the two wires, the mass should be suspended at a distance of 0.432 m from the end where wire A is attached

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Answered on 20/09/2018 Learn CBSE/Class 11/Science/Physics/Unit 7-Properties of Bulk Matter/Chapter 9-Mechanical Properties of Solids

Radius of the steel cable, r = 1.5 cm = 0.015 mMaximum allowable stress = 108 N m–2Maximum stress = Maximum force / Area of cross-section∴ Maximum force = Maximum stress × Area of cross-section= 108 × π (0.015)2= 7.065 × 104 NHence, the cable can support the maximum... ...more

Radius of the steel cable, r = 1.5 cm = 0.015 m
Maximum allowable stress = 108 N m–2
Maximum stress = Maximum force / Area of cross-section
∴ Maximum force = Maximum stress × Area of cross-section
= 108 × π (0.015)2
= 7.065 × 104 N
Hence, the cable can support the maximum load of 7.065 × 104 N.

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Answered on 20/09/2018 Learn CBSE/Class 11/Science/Physics/Unit 6-Gravitation/Chapter 8-Gravitation

Answer:- Mass of our galaxy Milky Way, M = 2.5 × 1011 solar massSolar mass = Mass of Sun = 2.0 × 1036 kgMass of our galaxy, M = 2.5 × 1011 × 2 × 1036 = 5 × 1041 kgDiameter of Milky Way, d = 105 lyRadius of Milky Way, r = 5 × 104 ly1 ly = 9.46 × 1015... ...more
Answer:-
 

Mass of our galaxy Milky Way, M = 2.5 × 1011 solar mass
Solar mass = Mass of Sun = 2.0 × 1036 kg
Mass of our galaxy, M = 2.5 × 1011 × 2 × 1036 = 5 × 1041 kg
Diameter of Milky Way, d = 105 ly
Radius of Milky Way, r = 5 × 104 ly
1 ly = 9.46 × 1015 m
r = 5 × 104 × 9.46 × 1015
= 4.73 ×1020 m
Since a star revolves around the galactic centre of the Milky Way, its time period is given by the relation:
= ( 4π2r3 / GM)1/2
= [ (4 × 3.142 × 4.733 × 1060) / (6.67 × 10-11 × 5 × 1041) ]1/2
=   (39.48 × 105.82 × 1030 / 33.35 )1/2
= 1.12 × 1016 s
1 year = 365 × 324 × 60 × 60 s
1s = 1 / (365 × 324 × 60 × 60)  years
∴ 1.12 × 1016 s = 1.12 × 1016 / (365 × 24 × 60 × 60)  =  3.55 × 108 year

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Answered on 20/09/2018 Learn CBSE/Class 11/Science/Physics/Unit 3-Laws of Motion/Chapter 5-Laws of Motion

Mass of the gun, M = 100 kg Mass of the shell, m = 0.020 kgMuzzle speed of the shell, v = 80 m/sRecoil speed of the gun = VBoth the gun and the shell are at rest initially.Initial momentum of the system = 0Final momentum of the system = mv – MVHere, the negative sign appears because the directions... ...more

Mass of the gun, M = 100 kg

Mass of the shell, m = 0.020 kg
Muzzle speed of the shell, v = 80 m/s
Recoil speed of the gun = V
Both the gun and the shell are at rest initially.
Initial momentum of the system = 0
Final momentum of the system = mv – MV
Here, the negative sign appears because the directions of the shell and the gun are opposite to each other.
According to the law of conservation of momentum:
Final momentum = Initial momentum
mv – MV = 0
∴ V = mv / M
= 0.020 × 80 / (100 × 1000) = 0.016 m/s

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Teaches

Engineering Diploma Tuition

Class Location

Online Classes (Video Call via UrbanPro LIVE)

Student's Home

Tutor's Home

Years of Experience in Engineering Diploma Tuition

5

Engineering Diploma Branch

Engineering Diploma 1st Year, Automobile Engineering Diploma, Mechanical Engineering Diploma

Engineering Diploma Subject

Basic Math, Engineering Mechanics, Basic Physics, Engineering Graphics, Engineering Mathematics

Mechanical Engineering Diploma Subject

Applied Mathematics, Design of Machine Elements, Thermal Engineering, Mechanical Engineering Materials, Fundamentals of Electronics, Mechanical Engineering Drawing, Strength of Materials, Power Plant Engineering, Automobile Engineering

Automobile Engineering Diploma Subject

Automobile Manufacturing Processes, Automobile Air Conditioning, Automobile component Design, Automotive Electrical and Electronic Syaytems, Applied Math, Strength of Materials, Mechanical Engineering Drawing, Vehicle aerodynamics and Design, Materials and Manufacturing Process, Heat Power Engineering, Advanced Automobile Engines, Vehicle Maintainence, Vehicle Testing, Automobile Systems, Automibile Engines, Automobile Transmission Systems

Type of class

Regular Classes

Class strength catered to

One on one/ Private Tutions

Taught in School or College

Yes

BTech Tuition
1 Student

Class Location

Online Classes (Video Call via UrbanPro LIVE)

Student's Home

Tutor's Home

Years of Experience in BTech Tuition

4

BTech Mechanical subjects

Internal Combustion Engines and Emissions, Turbomachines, Automobile Engineering, Fluid Mechanics, Engineering Drawing & Graphics, Automotive Fuels and Fuel Sysytems, Refrigeration & Air Conditioning Systems, Strength of Materials, Finite Element Analysis, Thermodynamics, Heat & Mass Transfer

BTech Branch

BTech Automobile Engineering, BTech 1st Year Engineering, BTech Mechantronics Engineering, BTech Mechanical Engineering

BTech Automobile subjects

New Generation And Hybrid Vehicles, Engineering Mechanics, Vehicle Dynamics, Autotronics, Automotive Pollution Control, Thermodynamics, Materials Science And Metallurgy, Strength Of Materials, Computational Fluid Dynamics, Automotive Engines, Engine Auxiliary Systems, Composite Materials And Mechanics, Automotive Chassis And Suspension, Engine Tribology, Two And Three Wheelers, Theory Of Machines, Ergonomics In Automotive Design, Alternative Fuels And Energy System, Automotive Electrical And Electronics System, Vehicle Aerodynamics, Automotive Safety

Type of class

Regular Classes

Class strength catered to

One on one/ Private Tutions

BTech Mechantronics subjects

Strength of Materials, Autotronics, Hybrid and Electric Vehicles, Automobile Engineering, Mechanical Vibrations, Design of Machine Elements, Engineering Thermodynamics and Heat Transfer

Taught in School or College

Yes

BTech 1st Year subjects

Engineering Physics, Basic Electronics, Engineering Graphics, Mechanics Of Solids, Basic Mechanical Engineering

Class 9 Tuition

Class Location

Online Classes (Video Call via UrbanPro LIVE)

Student's Home

Tutor's Home

Years of Experience in Class 9 Tuition

4

Board

CBSE, State

Subjects taught

Mathematics, Science, Tamil

Taught in School or College

Yes

Class 10 Tuition

Class Location

Online Classes (Video Call via UrbanPro LIVE)

Student's Home

Tutor's Home

Years of Experience in Class 10 Tuition

4

Board

CBSE, State

Subjects taught

Mathematics, Tamil, Science

Taught in School or College

Yes

Class 11 Tuition

Class Location

Online Classes (Video Call via UrbanPro LIVE)

Student's Home

Tutor's Home

Years of Experience in Class 11 Tuition

4

Board

State, CBSE

Subjects taught

Mathematics, English, Physics, Home Science, Tamil, Electronics

Taught in School or College

Yes

Teaching Experience in detail in Class 11 Tuition

4yrs

Class 12 Tuition

Class Location

Online Classes (Video Call via UrbanPro LIVE)

Student's Home

Tutor's Home

Years of Experience in Class 12 Tuition

4

Board

State, CBSE

Subjects taught

Tamil, Physics, Mathematics, Home Science, English, Electronics

Taught in School or College

Yes

Teaching Experience in detail in Class 12 Tuition

4yrs

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Answers by Balakumar (5)

Answered on 20/09/2018 Learn CBSE/Class 11/Science/Physics/Unit 7-Properties of Bulk Matter/Chapter 9-Mechanical Properties of Solids

Water pressure at the bottom, p = 1.1 × 108 PaInitial volume of the steel ball, V = 0.32 m3Bulk modulus of steel, B = 1.6 × 1011 Nm–2The ball falls at the bottom of the Pacific Ocean, which is 11 km beneath the surface.Let the change in the volume of the ball on reaching the bottom... ...more

Water pressure at the bottom, = 1.1 × 108 Pa
Initial volume of the steel ball, V = 0.32 m3
Bulk modulus of steel, B = 1.6 × 1011 Nm–2
The ball falls at the bottom of the Pacific Ocean, which is 11 km beneath the surface.
Let the change in the volume of the ball on reaching the bottom of the trench be ΔV.
Bulk modulus, B = p / (?V/V)
?V  =  pV/B
= 1.1 × 108 × 0.32 / (1.6 × 1011 )  =  2.2 × 10-4 m3
Therefore, the change in volume of the ball on reaching the bottom of the trench is 2.2 × 10–4 m3.

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Answered on 20/09/2018 Learn CBSE/Class 11/Science/Physics/Unit 7-Properties of Bulk Matter/Chapter 9-Mechanical Properties of Solids

Cross-sectional area of wire A, a1 = 1.0 mm2 = 1.0 × 10–6 m2Cross-sectional area of wire B, a2 = 2.0 mm2 = 2.0 × 10–6 m2Young’s modulus for steel, Y1 = 2 × 1011 Nm–2Young’s modulus for aluminium, Y2 = 7.0 ×1010Nm–2(a) Let a small mass m be suspended... ...more

Cross-sectional area of wire Aa1 = 1.0 mm= 1.0 × 10–6 m2
Cross-sectional area of wire Ba2 = 2.0 mm= 2.0 × 10–6 m2
Young’s modulus for steel, Y1 = 2 × 1011 Nm–2
Young’s modulus for aluminium, Y2 = 7.0 ×1010Nm–2

(a) Let a small mass m be suspended to the rod at a distance y from the end where wire A is attached.
Stress in the wire = Force / Area  =  F / a
If the two wires have equal stresses, then:
F1 / a1  =  F2 / a2
Where,
F1 = Force exerted on the steel wire
F2 = Force exerted on the aluminum wire
F1 / F2 = a1 / a2  =  1 / 2    ….(i)
The situation is shown in the following figure:

 
 

Taking torque about the point of suspension, we have:
F1y = F2 (1.05 – y)
F1 / F2 = (1.05 – y) / y    ……(ii)
Using equations (i) and (ii), we can write:
(1.05 – y) / y  = 1 / 2
2(1.05 – y)  =  y
y = 0.7 m
In order to produce an equal stress in the two wires, the mass should be suspended at a distance of 0.7 m from the end where wire A is attached.

(b) Young’s modulus = Stress / Strain
Strain = Stress / Young’s modulus  =  (F/a) / Y
If the strain in the two wires is equal, then:
(F1/a1) / Y1  =  (F2/a2) / Y2
F1 / F2 = a1Y1 / a2Y2
a1 / a2 = 1/2
F1 / F2 = (1 / 2) (2 × 1011 / 7 × 1010)  =  10 / 7      …….(iii)
Taking torque about the point where mass m, is suspended at a distance y1 from the side where wire A attached, we get:
F1yF2 (1.05 – y1)
F1 / F2  =  (1.05 – y1) / y1    ….(iii)
Using equations (iii) and (iv), we get:
(1.05 – y1) / y1  =  10 / 7
7(1.05 – y1)  =  10y1
y1 = 0.432 m
In order to produce an equal strain in the two wires, the mass should be suspended at a distance of 0.432 m from the end where wire A is attached

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Answered on 20/09/2018 Learn CBSE/Class 11/Science/Physics/Unit 7-Properties of Bulk Matter/Chapter 9-Mechanical Properties of Solids

Radius of the steel cable, r = 1.5 cm = 0.015 mMaximum allowable stress = 108 N m–2Maximum stress = Maximum force / Area of cross-section∴ Maximum force = Maximum stress × Area of cross-section= 108 × π (0.015)2= 7.065 × 104 NHence, the cable can support the maximum... ...more

Radius of the steel cable, r = 1.5 cm = 0.015 m
Maximum allowable stress = 108 N m–2
Maximum stress = Maximum force / Area of cross-section
∴ Maximum force = Maximum stress × Area of cross-section
= 108 × π (0.015)2
= 7.065 × 104 N
Hence, the cable can support the maximum load of 7.065 × 104 N.

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Answered on 20/09/2018 Learn CBSE/Class 11/Science/Physics/Unit 6-Gravitation/Chapter 8-Gravitation

Answer:- Mass of our galaxy Milky Way, M = 2.5 × 1011 solar massSolar mass = Mass of Sun = 2.0 × 1036 kgMass of our galaxy, M = 2.5 × 1011 × 2 × 1036 = 5 × 1041 kgDiameter of Milky Way, d = 105 lyRadius of Milky Way, r = 5 × 104 ly1 ly = 9.46 × 1015... ...more
Answer:-
 

Mass of our galaxy Milky Way, M = 2.5 × 1011 solar mass
Solar mass = Mass of Sun = 2.0 × 1036 kg
Mass of our galaxy, M = 2.5 × 1011 × 2 × 1036 = 5 × 1041 kg
Diameter of Milky Way, d = 105 ly
Radius of Milky Way, r = 5 × 104 ly
1 ly = 9.46 × 1015 m
r = 5 × 104 × 9.46 × 1015
= 4.73 ×1020 m
Since a star revolves around the galactic centre of the Milky Way, its time period is given by the relation:
= ( 4π2r3 / GM)1/2
= [ (4 × 3.142 × 4.733 × 1060) / (6.67 × 10-11 × 5 × 1041) ]1/2
=   (39.48 × 105.82 × 1030 / 33.35 )1/2
= 1.12 × 1016 s
1 year = 365 × 324 × 60 × 60 s
1s = 1 / (365 × 324 × 60 × 60)  years
∴ 1.12 × 1016 s = 1.12 × 1016 / (365 × 24 × 60 × 60)  =  3.55 × 108 year

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Answered on 20/09/2018 Learn CBSE/Class 11/Science/Physics/Unit 3-Laws of Motion/Chapter 5-Laws of Motion

Mass of the gun, M = 100 kg Mass of the shell, m = 0.020 kgMuzzle speed of the shell, v = 80 m/sRecoil speed of the gun = VBoth the gun and the shell are at rest initially.Initial momentum of the system = 0Final momentum of the system = mv – MVHere, the negative sign appears because the directions... ...more

Mass of the gun, M = 100 kg

Mass of the shell, m = 0.020 kg
Muzzle speed of the shell, v = 80 m/s
Recoil speed of the gun = V
Both the gun and the shell are at rest initially.
Initial momentum of the system = 0
Final momentum of the system = mv – MV
Here, the negative sign appears because the directions of the shell and the gun are opposite to each other.
According to the law of conservation of momentum:
Final momentum = Initial momentum
mv – MV = 0
∴ V = mv / M
= 0.020 × 80 / (100 × 1000) = 0.016 m/s

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Balakumar describes himself as Automobile Engineer. He conducts classes in BTech Tuition, Class 10 Tuition and Class 11 Tuition. Balakumar is located in Tambaram West, Chennai. Balakumar takes at students Home and Regular Classes- at his Home. He has 5 years of teaching experience . He is well versed in English and Tamil.

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