Hey,

In order to find the magnitude of a vector the formula is,

R^2= P^2 + Q^2 + 2 PQ cos¢

Now given, R= P+ Q

Squaring both sides,  R^2= P^2 + Q^2 + 2PQ

Equating both the reaction,

R^2= P^2 + Q^2 + 2 PQ cos¢   = R^2= P^2 + Q^2 + 2PQ

Simplifying, cos¢ = 1

= Cos 0° = 1

Therefore the angle is 0 degree.

Hope that helps.

Thank you.

Hey,

The answer of these type of questions are difficult to answer at this platform. You can get the answer to these type of question by referring to some question answering platform specifically decided for Ed related content.

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Hope that helps.

Thank you

Dear Student,

This question is concept based please refer your NCERT textbook.You should always learn from your textbook first then you can search online for solutions if you have doubt better learn from YouTube channel or join coaching institutes.

Thanks

Here lets assume m1 = 6kg and the corresponding position vector be r1=6i-7j and for m2 = 2 kg and corresponding position vector r2=2i+10j.

Now CM = m1r1+m2r2/(m1+m2)=(20/7) i - (11/7) j

Thus the center of mass will be located in the coordinate (20/7,-11/7)

I hope this answer helps. All the very best for your future endeavors!

Dear Student,

• In SHM, the velocity is maximum at the equilibrium position
• Velocity is zero at the extreme position
• Further if the velocity is maximum, the acceleration is zero, meaning that the acceleration is zero whenever the object is at the initial position or the displacement of the particle is zero
• Acceleration is maximum at the extreme position as the displacement is maximum at the extreme position

Dear Student,

• Work is defined as the product of work and displacement.
• Hence, the work done by force is found by finding the area under the force displacement graph.

Given, Radius of Curvature = R

Let, Static Friction = U

Now the centripetal force is providing the friction between the tyre and the road i.e.,

mv^2/R = Umg

=> v = (URg)^1/2

• Thus the Maximum speed will be independent of Mass.

I hope this answer helps. All the very best for your future endeavors!

Hello candidate,

For small oscillations, of a pendulum the angle taken between the two bobs were taken to be sin (theta), but we take the value of sin theta equal to theta. So, the expression for the time period of small oscillations is equal to 2π× √(length of the pendulum/ gravity) which is independent of the masses of the Bob attached.

Hope that this information was helpful for you!!

The various modes of vibration of air column are:

The air at the closed end is not free to vibrate so a node is formed at the closed end. While the air at the open end is free to vibrate with maximum amplitude so an antinode is formed at this end.The simplist mode is this one and it iscalled the fundamental mode.

Hope it helps!!