A particle is projected with velocity V along x-axis. A damping force is acting on the particle which is proportional to the square of the distance from the origin. i.e, ma = -x 2 . The distance at which the particle stops
Hello candidate,
In order to solve this problem you have to take the help of differential form of acceleration in the form of velocity-
We know acceleration of the particle is equal to- a= -2x/m, and the formula acceleration is equal to vdv/dx=-2x/m, so the value of v equals to x/√m.
Hope you found it helpful!!
For more queries, feel free to post it here!!
when the force acting onthe object and the velocity of the object both are along the same line, it is called...
Dear leaner,
when object and velocity is in same line .it is said to be according to Newton's first law of motion it is the basic one called linear motion .
Objects stays in rest until the external forces acted upon
Example for this is screwball. Hope you got it
All the best.
what are the acting schools in India that provide placement after course?
Hello candidate,
First of all, I want to clear that there are no acting schools which provide 100% placement record for its students because the field of acting is purely dependent on the candidates portfolio and his potential to showcase his skills on the screen when required. But, here are the name of a few top acting schools which might be of your interest-
Anupam Kher's Actor Prepares - Mumbai
Film and Television Institute of India - Pune
Satyajit Ray Film and Television Institute - Kolkata
Barry John Acting Studio - Mumbai
National School of Drama - New Delhi
Asian Academy of Film and Television - Mumbai, Kolkata, Noida and New Delhi
Good Luck!!
force system acting on particle can produce..options are...translation, rotational, deformation
Depending on the nature of the force system you can conclude the impact on the particle. So the answer can be all of the above.
I hope this answer helps. All the very best for your future endeavors!
derive an expression for the torque acting on the rectangular current carrying coil of a galvanometer
Dear Student,
F 1 = F 2 = BbI ,
As, these forces are equal and opposite, the net force is 0. Now, as theta component tends to 0, the perpendicular distance between the force of the coil also approaches 0. This makes the forces collinear and, the net force and torque becomes 0.
For a system of particles under Central force field, the total angular momentum is conserved. Reason (R): The torque acting on such a system is zero.
Under the central force field, the force acts along the line that joins the bodies so it has no rotary effect, that is, the torque is zero. Therefore, the angular momentum is conservative.
Both the assertion and reason are correct.
I hope this answer helps. All the very best for your future endeavors!
Assertion(A): For a system of particles under Central force field, the total angular momentum is conserved. Reason (R): The torque acting on such a system is zero.
Dear Student,
This type of question is just simple theoretical concept based you should study from books and read them because you will be here find the solution but you will not able to remember it.This assertion is true and you need to prove it.
Thanks
a force for=10N acting on a particle at an angle of 30 degrees with y-axis. find the horizontal and verticle components of force and then express force in terms of i and j
Deepika,
The solution of the query asked is given below:
As the force is making 30 degree angle with the y-axis so , for the x-component of force: Sin of 30 degree * force
= 10N Sin(30) = 5 N
X-component is 5N
For Y-component : 10*cos(30) = 8.66N
So, Fx = 5 N and Fy = 8.66N
Now, as we know: F = Xi + Yj ; where X = 5N and Y = 8.66N
Therefore, F = 5i + 8.66j
Hope it helps to clear your dilemma, thankyou!
The torque of force F = 2i -3j +4k newton acting at the point r =3i +2k +3k metre about origin is (in N-m) a).6i-6j +12k b).17i - 6j -13k c).-6i +6k -12k d).-17i +6j +13k
Dear aspirant,
For this question you just simply need to apply the formula for torque which is r×F where r = (3-0)i +(2-0)j+(3-0)k and F=(2i-3j+4k) , just do the normal cross product calculations which will give you 17i -6j-13k. This is your final torque expression.