Wagholi, Pune, India - 412207.
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NIT Silchar 2017
Bachelor of Technology (B.Tech.)
Wagholi, Pune, India - 412207
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Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Engineering Entrance Coaching classes
2
Engineering Entrance Exams
BITSAT Coaching Classes, Delhi CEE Coaching Classes, IIT JEE Coaching Classes
IITJEE Coaching
IIT JEE Mains Coaching, IIT JEE Foundation Course, IIT JEE Integrated Coaching, IIT JEE Crash Course
Type of class
Crash Course, Regular Classes
IIT-JEE Subjects
Maths
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 11 Tuition
2
Board
State, CBSE, ISC/ICSE
ISC/ICSE Subjects taught
Mathematics
CBSE Subjects taught
Mathematics
Taught in School or College
Yes
State Syllabus Subjects taught
Mathematics
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 12 Tuition
2
Board
State, CBSE, ISC/ICSE
ISC/ICSE Subjects taught
Mathematics
CBSE Subjects taught
Mathematics
Taught in School or College
Yes
State Syllabus Subjects taught
Mathematics
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in BTech Tuition
1
BTech Mechanical subjects
Strength of Materials, Engineering Drawing & Graphics, Material Science and Metallurgy, Dynamics of Machinery, Automobile Engineering, Manufacturing Technology, Heat & Mass Transfer, Kinematics of Machinery, Analysis and Design of Machine Components, Machine Design, Mechanics of Machines, Fluid Mechanics, Internal Combustion Engines and Emissions
BTech Branch
BTech Mechanical Engineering
Type of class
Crash Course, Regular Classes
Class strength catered to
One on one/ Private Tutions, Group Classes
Taught in School or College
Yes
1. Which classes do you teach?
I teach BTech Tuition, Class 11 Tuition, Class 12 Tuition and Engineering Entrance Coaching Classes.
2. Do you provide a demo class?
Yes, I provide a free demo class.
3. How many years of experience do you have?
I have been teaching for 2 years.
Answered on 09/01/2018 Learn Tuition/Class XI-XII Tuition (PUC)
Sec(360)=1/cos(360).
If we take a stick of length l and fix one of its end to the origin. It is free to rotate from its another end and will make circle at Cartesian coordinate plane. Rotate it by 90º it will lie on positive y axis. Similarly rotate it by 180° it will lie on negative x axis. If we rotate it by 360° we will find that it came in its original position i.e lies on positive x axis. Imagine a position where stick was just about complete. Take the projection of stick on x axis. We get a right angle triangle in third quadrant. Calling the angle between positive x axis and stick as θ. Now cosθ=(projection of stick on positive x axis) / (length of stick=l).
As θ decreases. Projection of stick will increase and when stick completes its 360° rotation it will be equal to l.
At that time
Cosθ = l/l =1.
⇒secθ = 1/cosθ =1/1 =1 Ans.
Answered on 09/01/2018 Learn CBSE/Class 11/Science/Physics
Total time taken by the particle to go up and come back = 5+9 =14s.
Therefore it must take 7s to go up and 7s to come down.
From first law, we know
v=u+at
At the top point velocity will be zero. Therefore v=0.
0=u-10*7
(taking a=g=-10m/s², negative sign indicates that acceleration is in downward direction). Solving the equation we will get
u=70m/s.
It is given that after 5s its height was h.
Again applying first law.
v=u+at
v=70-10*5
v= 20m/s
Particle speed was 20m/s in upward direction at height h after 5s of throwing.
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Engineering Entrance Coaching classes
2
Engineering Entrance Exams
BITSAT Coaching Classes, Delhi CEE Coaching Classes, IIT JEE Coaching Classes
IITJEE Coaching
IIT JEE Mains Coaching, IIT JEE Foundation Course, IIT JEE Integrated Coaching, IIT JEE Crash Course
Type of class
Crash Course, Regular Classes
IIT-JEE Subjects
Maths
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 11 Tuition
2
Board
State, CBSE, ISC/ICSE
ISC/ICSE Subjects taught
Mathematics
CBSE Subjects taught
Mathematics
Taught in School or College
Yes
State Syllabus Subjects taught
Mathematics
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 12 Tuition
2
Board
State, CBSE, ISC/ICSE
ISC/ICSE Subjects taught
Mathematics
CBSE Subjects taught
Mathematics
Taught in School or College
Yes
State Syllabus Subjects taught
Mathematics
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in BTech Tuition
1
BTech Mechanical subjects
Strength of Materials, Engineering Drawing & Graphics, Material Science and Metallurgy, Dynamics of Machinery, Automobile Engineering, Manufacturing Technology, Heat & Mass Transfer, Kinematics of Machinery, Analysis and Design of Machine Components, Machine Design, Mechanics of Machines, Fluid Mechanics, Internal Combustion Engines and Emissions
BTech Branch
BTech Mechanical Engineering
Type of class
Crash Course, Regular Classes
Class strength catered to
One on one/ Private Tutions, Group Classes
Taught in School or College
Yes
Answered on 09/01/2018 Learn Tuition/Class XI-XII Tuition (PUC)
Sec(360)=1/cos(360).
If we take a stick of length l and fix one of its end to the origin. It is free to rotate from its another end and will make circle at Cartesian coordinate plane. Rotate it by 90º it will lie on positive y axis. Similarly rotate it by 180° it will lie on negative x axis. If we rotate it by 360° we will find that it came in its original position i.e lies on positive x axis. Imagine a position where stick was just about complete. Take the projection of stick on x axis. We get a right angle triangle in third quadrant. Calling the angle between positive x axis and stick as θ. Now cosθ=(projection of stick on positive x axis) / (length of stick=l).
As θ decreases. Projection of stick will increase and when stick completes its 360° rotation it will be equal to l.
At that time
Cosθ = l/l =1.
⇒secθ = 1/cosθ =1/1 =1 Ans.
Answered on 09/01/2018 Learn CBSE/Class 11/Science/Physics
Total time taken by the particle to go up and come back = 5+9 =14s.
Therefore it must take 7s to go up and 7s to come down.
From first law, we know
v=u+at
At the top point velocity will be zero. Therefore v=0.
0=u-10*7
(taking a=g=-10m/s², negative sign indicates that acceleration is in downward direction). Solving the equation we will get
u=70m/s.
It is given that after 5s its height was h.
Again applying first law.
v=u+at
v=70-10*5
v= 20m/s
Particle speed was 20m/s in upward direction at height h after 5s of throwing.
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