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Syllabus

A particle is moving in a circle of radius R with constant speed . The time period of particle is T=1 Second in a time t = T/6 , If the difference between average speed and magnitude of average velocity of the partical is 2m/sec, find the radius of the circle(in meteres)

A)The velocity vector is tangential to the circle

B)The acceleration vector is tangetial to the circle

C)The acceleration vector is directed towards the centre of the circle

D)The velocity and acceleration vectors are perpendicular to each other

^{3}+21t^{2}+60t+6 where X is in meter And t is in second . find the acceleration of the particle when its velocity is zero ?^{2}. If the initial velocity is u, the distance covered in 't' seconds isa) aut

b) 1/a ln(aut)

c) 1/a ln(1+aut)

d) aln(aut)

2 particles P and Q are moving with constant velocities ( i + j ) m/s and (-i + 2j) m/s respectively.At time t=0, P is at origin and Q is at a point with position vector (2i + j) m. If the equation of the trajectory of Q as observed by P is x+2y=n find n^{2}. After 5 s, a ball is dropped through window of car. The window of the car is at a height of 1.5m from the ground. What will be the speed of ball 0.5 s after it was dropped ?a) 5 m/s

b) (5)

^{1/2}m/sc) 5(5)

^{1/2}m/sd) 2 m/s

velocityAccelerationa) +ve +ve

b) +ve -ve

c) +ve 0

d)-ve +ve

e)-ve -ve

f) -ve 0

g)0 +ve

h) 0 -ve

Describe what the particle is doing in each case, and give a real life example for an automobile on an east-west one-dimensional axis, with east considered to be the +ve direction.

what I have to do?

The coordinate of point B is

please answer as soon as possible!!

speed" change in a uniform circular motion ........if yes ??? i think it is scalar and change in direction wont make any change

a) The velocity of the particle is constant

b) The acceleration of the particle is constant

c) The magnitude of acceleration is constant.

c) The magnitude of acceleration s decreasing continuously.

^{0 }with the horizontal with an initial speed of 20 m/sec, with H being highest point of its trajectory another particle Pforced to move along the same trajectory as that of the projectile such that its speed is continuously increasing when the particle P is at H, mod vector vp = 20 m/sec , mod vector ap -= 50 m/sec^{is now }^{2, ,}then acceleartion vector ap, at H equals(take g = 10 m/sec^{2}^{-1}.Q: The position vector of a particle in a circular motion about the origin sweeps out equal area in equal time is ?

{Take the horizontal as the reference level for the gravitational potential energy.}

(1)

(2)

(3)

(4)

^{3/2}. If the particle starts its motion fro origin with velocity of 4m/s, the position x of the particle at an instant terms of v can be given as^{}A SWIMMER SWIMS ACROSS A 2 KM WIDE RIVER BY KEEPING 150 DEGREE WITH FLOW OF RIVER. THE RIVER FLOWS WITH 2KM/H AND THE SPEED OF THE SWIMMER IN STILL WATER IS 3KM/H. FIND THE TIME HE TAKES TO CROSS THE RIVER.PLS XPLAIN IN DETAIL WITH A DIAGRAM

a person standing on the roof of a 40 m high tower, throws a ball vertically upwards with speed 10m/s. 2 seconds later, he throws another ball again in vertical downward direction(g=10m/s^2) if both hit the ground simultaneously, then choose the correct options.a) the first ball hits the ground after 4 seconds.

b) the second ball was projected vertically downwards with speed 5m/s.

c) the distance travelled by the first ball is 10 m greater than the distance travelled by the second ball.

d) both balls hit the ground with same velocities.

^{2}for 4.2s, making the straight skid marks 62.4m long ending at the tree. With what speed does the car then strike the tree?t= 0,x= +10 m. In which region particle is moving towards the origin?Options:ABandBCBCandDEBCandCDCDandDET=sqrt(2Rtantheta/g) tantheta=gT^2/2R

A hemispherical bowl of radius 10 m rotates about its vertical axis with an angular velocity 1.4 rad/s. A point particle of mass 10-2 kg placed on the smooth inner surface of the bowl, also rotates with the same speed. The particle is at a height h from the bottom of the bowl. Find the height of the particle.

^{2}-4s)/a].^{-2}. Then find the maximum distance between A and B before B overtakes Aif acceleration is varying with time we does not use kinematics equation,.we will use this equation for constant acceleration why plzexplain?