Labib J.

Wanowrie, Pune, India - 411040

Wanowrie, Pune, India - 411040.

3.9

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Labib J. describes himself as Progressive classes.. He conducts classes in Class 10 Tuition, Class 9 Tuition and Engineering Diploma Tuition. Labib is located in Wanowrie, Pune. Labib takes at students Home and Regular Classes- at his Home. He has 1 years of teaching experience . Labib is pursuing Master of Engineering - Master of Technology (M.E./M.Tech.) from Sspu. Labib has completed Bachelor of Engineering (B.E.) from Sspu in 2011. He is well versed in Marathi, Urdu, Hindi and English.

Marathi

Urdu

Hindi

English

Sspu Pursuing

Master of Engineering - Master of Technology (M.E./M.Tech.)

Sspu 2011

Bachelor of Engineering (B.E.)

Wanowrie, Pune, India - 411040

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Class 9 Tuition

Class Location

Student's Home

Tutor's Home

Online (video chat via skype, google hangout etc)

Years of Experience in Class 9 Tuition

1

Board

State

Taught in School or College

No

State Syllabus Subjects taught

Science, Mathematics

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Answered on 09/12/2017 CBSE/Class 9/Science Tuition/Class IX-X Tuition

Derive the formula for potential energy.

Consider body of mass is raised to a height 'h' above the ground level, the force acting on the body is the gravitational of the earth and equal to mg. This force acts in the vertically downwarddirection.To lift the body above the surface of tge earth, we will have to do work against this force of gravity. Let ... ...more

Consider body of mass is raised to a height 'h' above the ground level, the force acting on the body is the gravitational of the earth and equal to mg. This force acts in the vertically downwarddirection.To lift the body above the surface of tge earth, we will have to do work against this force of gravity. Let Work= force×displacement W = m×g×h W=mgh This work done on the body is actually the increase in potential energy of tge body. Hence work done is equal to potential energy. PE = mgh.

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Answers 1 Comments Answered on 08/12/2017 CBSE/Class 9/Science Tuition/Class IX-X Tuition

Mass number of Helium = 4 Number of protons = 2 Number of neutrons(n) = Number of Mass(A) Number of proton (P) = 4 -- 2 = 2 Thus, No. of neutrons = 2

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Answers 9 Comments Answered on 07/12/2017 CBSE/Class 10/Science Tuition/Class IX-X Tuition

Let L= length of wire d=daimeter of wire R= resistance of wire in ohms Give data L= 1m d= 0.2mm converted mm into meter d=0.2 10^-3 m R= 10? R=p L/A Area=pi (d/2)^2 =3.14 (0.2 10^-3/2)^2 =3.1415 10^-08 R=p L/A R=pL/A 10=p (1/3.1415 10^-08) Resistivity =3.14 10^-07 ?m

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Answers 6 Comments Answered on 07/12/2017 CBSE/Class 9/Mathematics Tuition/Class IX-X Tuition

Mean will be 75

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Answers 49 Comments Answered on 07/12/2017 CBSE/Class 9/Mathematics Tuition/Class IX-X Tuition

Let consider 141x + 103y = 217.....equation (1) 103x + 141y=27..equation(2) Multiple eq(1) by 103 we get.. 14523x + 10609y=22352..eq(3) Similarly multiple eq(2) by 141 we get.. 14523x + 19881y=3807..eq(4) Now solve eq 3 nd 4 14523x +10609y=22351 14523x+19881y=3807 - ... ...more

Let consider 141x + 103y = 217.....equation (1) 103x + 141y=27..equation(2) Multiple eq(1) by 103 we get.. 14523x + 10609y=22352..eq(3) Similarly multiple eq(2) by 141 we get.. 14523x + 19881y=3807..eq(4) Now solve eq 3 nd 4 14523x +10609y=22351 14523x+19881y=3807 - - - ____________________ -9272y= 18544 Y= -2 Put y=-2 in eq (1) 141x + 103×(-2)=217 141x - 206=217 141x=206+217 141x=423 X=3 So we get X=3 & Y=-2

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Answers 16 Comments Labib J.Directions

x Class 9 Tuition 3.9

Class Location

Student's Home

Tutor's Home

Online (video chat via skype, google hangout etc)

Years of Experience in Class 9 Tuition

1

Board

State

Taught in School or College

No

State Syllabus Subjects taught

Science, Mathematics

Class 10 Tuition 3.9

Class Location

Student's Home

Tutor's Home

Online (video chat via skype, google hangout etc)

Years of Experience in Class 10 Tuition

1

Board

State

Taught in School or College

No

State Syllabus Subjects taught

Science, Mathematics

Engineering Diploma Tuition 3.9

Class Location

Student's Home

Tutor's Home

Online (video chat via skype, google hangout etc)

Engineering Diploma Branch

Automobile Engineering Diploma, Mechanical Engineering Diploma

Mechanical Engineering Diploma Subject

Applied Mathematics, Fundamentals of Electronics, Theory of Machines & Mechanisms, Mechatronics, Design of Machine Elements, Industrial Fluid Power, Power Plant Engineering, Strength of Materials, Power Engineering, Mechanical Engineering Drawing, Fluid Mechanics and Machinery

Automobile Engineering Diploma Subject

Applied Math, Theory of Machines & Mechanisms, Strength of Materials, Mechanical Engineering Drawing, Automobile Transmission Systems, Automobile component Design

Type of class

Regular Classes, Crash Course

Class strength catered to

One on one/ Private Tutions

Taught in School or College

No

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No Reviews yet! Be the first one to Review

Answered on 09/12/2017 CBSE/Class 9/Science Tuition/Class IX-X Tuition

Derive the formula for potential energy.

Consider body of mass is raised to a height 'h' above the ground level, the force acting on the body is the gravitational of the earth and equal to mg. This force acts in the vertically downwarddirection.To lift the body above the surface of tge earth, we will have to do work against this force of gravity. Let ... ...more

Consider body of mass is raised to a height 'h' above the ground level, the force acting on the body is the gravitational of the earth and equal to mg. This force acts in the vertically downwarddirection.To lift the body above the surface of tge earth, we will have to do work against this force of gravity. Let Work= force×displacement W = m×g×h W=mgh This work done on the body is actually the increase in potential energy of tge body. Hence work done is equal to potential energy. PE = mgh.

Like 0

Answers 1 Comments Answered on 08/12/2017 CBSE/Class 9/Science Tuition/Class IX-X Tuition

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Answers 9 Comments Answered on 07/12/2017 CBSE/Class 10/Science Tuition/Class IX-X Tuition

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Answers 6 Comments Answered on 07/12/2017 CBSE/Class 9/Mathematics Tuition/Class IX-X Tuition

Mean will be 75

Like 0

Answers 49 Comments Answered on 07/12/2017 CBSE/Class 9/Mathematics Tuition/Class IX-X Tuition

Let consider 141x + 103y = 217.....equation (1) 103x + 141y=27..equation(2) Multiple eq(1) by 103 we get.. 14523x + 10609y=22352..eq(3) Similarly multiple eq(2) by 141 we get.. 14523x + 19881y=3807..eq(4) Now solve eq 3 nd 4 14523x +10609y=22351 14523x+19881y=3807 - ... ...more

Let consider 141x + 103y = 217.....equation (1) 103x + 141y=27..equation(2) Multiple eq(1) by 103 we get.. 14523x + 10609y=22352..eq(3) Similarly multiple eq(2) by 141 we get.. 14523x + 19881y=3807..eq(4) Now solve eq 3 nd 4 14523x +10609y=22351 14523x+19881y=3807 - - - ____________________ -9272y= 18544 Y= -2 Put y=-2 in eq (1) 141x + 103×(-2)=217 141x - 206=217 141x=206+217 141x=423 X=3 So we get X=3 & Y=-2

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Answers 16 Comments X

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