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Yash Pal Singh Class 9 Tuition trainer in Delhi

Yash Pal Singh

East of Kailash, Delhi, India - 110048.

1 Student

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Referral Discount: Get ₹ 250 off when you make a payment to start classes. Get started by Booking a Demo.

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Overview

Yash Pal Singh conducts classes in Class 10 Tuition and Class 9 Tuition. Yash is located in East of Kailash, Delhi. Yash takes at students Home, Regular Classes- at his Home and Online Classes- via online medium. He has 5 years of teaching experience . Yash has completed Bachelor of Technology (B.Tech.) from GGSIPU. He is well versed in Hindi and English.

Languages Spoken

Hindi

English

Education

GGSIPU

Bachelor of Technology (B.Tech.)

Address

East of Kailash, Delhi, India - 110048

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Teaches

Class 9 Tuition
1 Student

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 9 Tuition

5

Board

CBSE, ICSE

CBSE Subjects taught

Science, Mathematics, Social science

ICSE Subjects taught

Chemistry, Mathematics, English, Physics

Taught in School or College

No

Class 10 Tuition
1 Student

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 10 Tuition

5

Board

CBSE, ICSE

CBSE Subjects taught

Science, Mathematics, Social science

ICSE Subjects taught

Chemistry, Mathematics, English, Physics

Taught in School or College

No

Reviews

No Reviews yet!

FAQs

1. Which school boards of Class 10 do you teach for?

CBSE, ICSE

2. Do you have any prior teaching experience?

No

3. Which classes do you teach?

I teach Class 10 Tuition and Class 9 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 Yash Pal Singh (2)

Answered on 26/08/2016 Learn Tuition/Class IX-X Tuition

Total Time = 2 min and 20 seconds = 2*60 +20 = 120+20 = 140 seconds Distance covered in one round =?D =?×200 Now total round can be completed in the given time = 140÷40 = 3.5 Distance = ?×200×3.5 = 22÷7×200×3.5 ... ...more
Total Time = 2 min and 20 seconds = 2*60 +20 = 120+20 = 140 seconds Distance covered in one round =?D =?×200 Now total round can be completed in the given time = 140÷40 = 3.5 Distance = ?×200×3.5 = 22÷7×200×3.5 =2200 m Displacement is shortest distance between initial and final point As the men has covered 3 final and 1 half round ,so the shortest distance is its diameter which is 200 m So final answer are Distance = 2200m Displacement= 200m
Answers 62 Comments
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Answered on 26/08/2016 Learn Tuition/Class IX-X Tuition

It's simple As we know Work = Force X Displacement W=F.D So if any of the above two quantities is zero i.e. force or displacement then work is ZERO. 1) If a person is standing at a position and not moving holding an object. Then the work done in this case is zero because he has not moved from his... ...more
It's simple As we know Work = Force X Displacement W=F.D So if any of the above two quantities is zero i.e. force or displacement then work is ZERO. 1) If a person is standing at a position and not moving holding an object. Then the work done in this case is zero because he has not moved from his initial position i.e. displacement is zero. 2) If body moved from its position without any application of force . then also work done is zero.
Answers 89 Comments
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Teaches

Class 9 Tuition
1 Student

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 9 Tuition

5

Board

CBSE, ICSE

CBSE Subjects taught

Science, Mathematics, Social science

ICSE Subjects taught

Chemistry, Mathematics, English, Physics

Taught in School or College

No

Class 10 Tuition
1 Student

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 10 Tuition

5

Board

CBSE, ICSE

CBSE Subjects taught

Science, Mathematics, Social science

ICSE Subjects taught

Chemistry, Mathematics, English, Physics

Taught in School or College

No

No Reviews yet!

Answers by Yash Pal Singh (2)

Answered on 26/08/2016 Learn Tuition/Class IX-X Tuition

Total Time = 2 min and 20 seconds = 2*60 +20 = 120+20 = 140 seconds Distance covered in one round =?D =?×200 Now total round can be completed in the given time = 140÷40 = 3.5 Distance = ?×200×3.5 = 22÷7×200×3.5 ... ...more
Total Time = 2 min and 20 seconds = 2*60 +20 = 120+20 = 140 seconds Distance covered in one round =?D =?×200 Now total round can be completed in the given time = 140÷40 = 3.5 Distance = ?×200×3.5 = 22÷7×200×3.5 =2200 m Displacement is shortest distance between initial and final point As the men has covered 3 final and 1 half round ,so the shortest distance is its diameter which is 200 m So final answer are Distance = 2200m Displacement= 200m
Answers 62 Comments
Dislike Bookmark

Answered on 26/08/2016 Learn Tuition/Class IX-X Tuition

It's simple As we know Work = Force X Displacement W=F.D So if any of the above two quantities is zero i.e. force or displacement then work is ZERO. 1) If a person is standing at a position and not moving holding an object. Then the work done in this case is zero because he has not moved from his... ...more
It's simple As we know Work = Force X Displacement W=F.D So if any of the above two quantities is zero i.e. force or displacement then work is ZERO. 1) If a person is standing at a position and not moving holding an object. Then the work done in this case is zero because he has not moved from his initial position i.e. displacement is zero. 2) If body moved from its position without any application of force . then also work done is zero.
Answers 89 Comments
Dislike Bookmark

Yash Pal Singh conducts classes in Class 10 Tuition and Class 9 Tuition. Yash is located in East of Kailash, Delhi. Yash takes at students Home, Regular Classes- at his Home and Online Classes- via online medium. He has 5 years of teaching experience . Yash has completed Bachelor of Technology (B.Tech.) from GGSIPU. He is well versed in Hindi and English.

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