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Lesson Posted on 17 Mar CBSE/Class 9 Suvam Banerjee

I am a B.Tech Graduate from the West Bengal University of Technology giving home tuition. I have more...

Highest Common Factor (HCF): It is also called the Greatest Common divisor. When the greatest number divides perfectly two or more given numbers, the number is called HCF. Least Common Multiple (LCM): The least number divisible by two or more given numbers; that least number is called LCM of the numbers. Factor... read more

Highest Common Factor (HCF): It is also called the Greatest Common divisor. When the greatest number divides perfectly two or more given numbers, the number is called HCF.

Least Common Multiple (LCM): The least number divisible by two or more given numbers; that least number is called LCM of the numbers.

Factor and Multiples: If a number m divides perfectly second number n, then m is called the factor of n and n is called multiple of m.

Rule 1 : 1st number * 2nd number = LCM*HCF

There are two methods for calculating HCF and LCM :

• Factor Method
• Division Method

If the ratio of two numbers is a:b, then numbers are ak and bk, where k is a constant and hence, HCF is k and LCM is abk.

Rule 2 : LCM of fractions = LCM of numerators / HCF of denominators

HCF of fractions = HCF of numerators / LCM of denominators

If there is no common factor between both numbers, then LCM will be the numbers' product.

When a number is divided by a,b or c, leaving remainders p,q or r respectively such that the difference between the divisor and reminder in each case is the same, i.e. (a-p)=(b-q)=(c-r) = t. The least number must be in the form of (k-t), where k is the LCM of a,b and c.

The largest numbers, which when divide a,b and c, give remainders p,q and r respectively, are given by HCF of (a-p), (b-q) and (c-r).

The largest number, which when dividing the numbers a,b and c the remainders are same then the largest number is given by HCF of (a-b), (b-c) and (c-a).

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Lesson Posted on 26/12/2020 Exam Coaching/Engineering Entrance Coaching/IIT JEE Coaching CBSE/Class 9/Science/Unit 3-Motion, Force and Work/Motion Exam Coaching/Engineering Entrance Coaching/IIT JEE Coaching/IIT JEE Foundation Course I started teaching in 2011 while pursuing a B.tech computer science course. Being hard-working and curious,...

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Lesson Posted on 26/12/2020 Exam Coaching/Engineering Entrance Coaching/IIT JEE Coaching/IIT JEE Foundation Course Exam Coaching/Engineering Entrance Coaching/IIT JEE Coaching CBSE/Class 9/Science/Unit 3-Motion, Force and Work/Force and laws of motion I started teaching in 2011 while pursuing a B.tech computer science course. Being hard-working and curious,...

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Lesson Posted on 18/11/2020 CBSE/Class 9 Tuition Hazel

Distance The total path covered by an object.Scalar quantitySI unit is mDisplacementShortest distance covered by an object.Vector quantitySI unit is mSpeedSpeed = distance/timei.e. Rate at which an object covers a distanceIt is a scalar quantity SI unit m/sVelocityVelocity = displacement/ timei.e. Rate... read more

Distance
The total path covered by an object.
Scalar quantity
SI unit is m
Displacement
Shortest distance covered by an object.
Vector quantity
SI unit is m
Speed
Speed = distance/time
i.e. Rate at which an object covers a distance
It is a scalar quantity
SI unit m/s
Velocity
Velocity = displacement/ time
i.e. Rate and direction of change in the position of an object
It is a vector quantity
SI unit is m/s
Acceleration
Rate of change of velocity of an object with respect to time.
Acceleration = change in velocity/time
It is a vector quantity
SI unit is m/s^2

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Lesson Posted on 09/09/2020 CBSE/Class 9/Social Studies/History I am an educator with 25 plus years of experience in teaching Social Science, Economics at various levels...

This History lesson is enormous, and children often end up being confused. They would have read the whole lesson but wouldn't know the topics they need to focus on. Hence mind maps would be required while learning such lessons. Here are a few examples of how I would go about teaching this lesson. It... read more

This History lesson is enormous, and children often end up being confused. They would have read the whole lesson but wouldn't know the topics they need to focus on. Hence mind maps would be required while learning such lessons. Here are a few examples of how I would go about teaching this lesson. It would be a combination of various teaching methodologies.

DAY 1 LESSON PLAN

Children should first understand what allied powers, central powers and Axis powers are:

 Allied Powers WW 1, WW2 UK, France, USSR, USA (entered late) Central powers WW1 Germany, Austria, Hungary, Turkey Axis powers WW2 Germany, Itlay, Japan

DATES TO BE NOTED

 World war I - 1914 - 1918 World war II - 1939 - 1945

OVERVIEW OF THE SEQUENCE OF EVENTS LEADING TO THE RISE OF HITLER

1. Germany (Imperial Germany) fought 1st world war with Allies

2. Allies lost lots of resources.

3. The USA joins partners & defeats Germany.

4. End of imperial rule in Germany

5. Formation of democratic construction with federal structure - Weimer Republic

6. The peace treaty of Versailles - with allies

7. Hyperinflation in Germany

8. Introduction of Dawes plan by the USA - bail out Germany

9. Depression of 1924 - 1928 (worldwide)

10. The USA withdraws support to Germany

11. The economic crisis in Germany

12. People lose confidence in a democratic parliamentary system

13. Rise of Hitler through the Nazi party

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Lesson Posted on 26/05/2019 Recently i am teaching student from Chirec School and in college i used to be a match lecturer while...

Hola Amigos,

So In this Video you will get to know the overview on how to prepare for maths and get 100 out of 100. Watch it out!

Dislike Bookmark Real education is to educate the minds of the children.

I would suggest offline coaching because it binds the students and gives different levels of class surrounding, which is not possible in online classes. However, for extra self-studies, one can take the help of online material.
Dislike Bookmark AN EXPERTISE INSTITUTE OF SCIENCE AND COMMERCE STREAM

100g of sodium because of less atomic mass.
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Lesson Posted on 09/04/2018 Soumi Roy

I am an experienced, qualified tutor with over 8 years of experience in teaching maths and Science across...

Type 1: Question 1: Calculate the force needed to speed up a car with a rate of 5ms–2, if the mass of the car is 1000 kg. Solution: According to question: Acceleration (a) = 5m/s2 and Mass (m) = 1000 kg, therefore, Force (F) =?We know that, F = m x a= 1000 kg x 5m/s2= 5000 kg m/s2Therefore,... read more

Type 1:

Question 1: Calculate the force needed to speed up a car with a rate of 5ms–2, if the mass of the car is 1000 kg.

Solution: According to question:

Acceleration (a) = 5m/s2 and Mass (m) = 1000 kg, therefore, Force (F) =?
We know that, F = m x a
= 1000 kg x 5m/s2
= 5000 kg m/s2
Therefore, required Force = 5000 m/s2 or 5000 N

Question 2: If the mass of a moving object is 50 kg, what force will be required to speed up the object at a rate of 2ms–2?

Solution: According to the question;

Acceleration (a) = 2ms–2 and Mass (m) = 50 kg, therefore, Force (F) =?
We know that, F = m x a
= 50 kg x 2m/s2
= 100 kg m/s2
Therefore, required Force = 100 m/s2 or 100 N

Question 3: To accelerate a vehicle to 3m/s2 what force will be needed if the mass of the vehicle is equal to 100 kg?

Solution: According to the question:

Acceleration (a) = 3m/s2 and Mass (m) = 100 kg, therefore, Force (F) =?
We know that, F = m x a
= 100 kg x 3m/s2
= 300 kg m/s2
Therefore, required Force = 300 m/s2 or 300 N

Type II:

Question 1: To accelerate an object to a rate of 2m/s2, 10 N force is required. Find the mass of object.

Solution: According to the question:

Acceleration (a) = 2m/s2, Force (F) = 10N, therefore, Mass (m) = ?
We know ,

F=m×a

10N=m×2m/s2

m=10/2 kg=5 kg

Thus, the mass of the object = 5 kg

Question 2: If 1000 N force is required to accelerate an object to the rate of 5m/s2, what will be the weight of the object?

Solution: According to the question,

Acceleration (a) = 2m/s2, Force (F) = 1000N, therefore, Mass (m) = ?
We know that,

F=m×a

1000N=m×5 m/s2

m=1000/5 kg=200 kg

Thus, the mass of the object = 200 kg

Question 3: A vehicle accelerate at the rate of 10m/s2 after the applying of force equal to 50000 N. Find the mass of the vehicle.

Solution: According to the question,

Acceleration (a) = 10 m/s2, Force (F) = 50000N, therefore, Mass (m) = ?
We know that,

F=m×a

50000 N=m×10m/s2

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Open-air evaporation on seashores forms concentrated brine which continues to form a supersaturated solution finally precipitation takes place, and salt get separated.
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