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Question Number 25509    Answers: 0   Comments: 1

Question Number 25736    Answers: 0   Comments: 0

Question Number 25483    Answers: 1   Comments: 0

valute ∫(1/((x−1)^2 ))(/((x^2 +4)))dx

valute1(x1)2(x2+4)dx

Question Number 25476    Answers: 1   Comments: 0

convert 1234.24_8 to base 10

convert1234.248tobase10

Question Number 25472    Answers: 1   Comments: 0

slolve the definite integral∫_1^ ^2 (1/x)dxusingg> ng trapezoidal rule with 4 sub intervals hencefind an approximate value of ln 2.

slolvethedefiniteintegral121xdxusingg>ngtrapezoidalrulewith4subintervalshencefindanapproximatevalueofln2.

Question Number 25414    Answers: 1   Comments: 0

if 10^(10 ) electrons are removed neutral bodyb body the charge acquired by the body is? ?

if1010electronsareremovedneutralbodybbodythechargeacquiredbythebodyis??

Question Number 25410    Answers: 0   Comments: 0

Question Number 25780    Answers: 1   Comments: 0

every periodic function is differentiable.true or false justify

everyperiodicfunctionisdifferentiable.trueorfalsejustify

Question Number 25275    Answers: 0   Comments: 0

Question Number 25103    Answers: 1   Comments: 0

Question Number 25091    Answers: 1   Comments: 0

A particle of mass m moving with speed u collides perfectly inelastically with a sphere of radius R and same mass, at rest, at an impact parameter d. Find (a) Angle between their final velocities (b) Magnitude of their final velocities

AparticleofmassmmovingwithspeeducollidesperfectlyinelasticallywithasphereofradiusRandsamemass,atrest,atanimpactparameterd.Find(a)Anglebetweentheirfinalvelocities(b)Magnitudeoftheirfinalvelocities

Question Number 25086    Answers: 0   Comments: 4

Question Number 25058    Answers: 1   Comments: 0

Two objects slide over a frictionless horizontal surface. The first object, mass m_1 = 5 kg, is propelled with a speed u = 4.5 m/s towards the second object, mass m_2 = 5 kg, which is initially at rest. After the collision, both objects have velocities which are directed at θ = 60° on either side of the original line of motion of the first object. What can you say about the elasticity of collision?

Twoobjectsslideoverafrictionlesshorizontalsurface.Thefirstobject,massm1=5kg,ispropelledwithaspeedu=4.5m/stowardsthesecondobject,massm2=5kg,whichisinitiallyatrest.Afterthecollision,bothobjectshavevelocitieswhicharedirectedatθ=60°oneithersideoftheoriginallineofmotionofthefirstobject.Whatcanyousayabouttheelasticityofcollision?

Question Number 25038    Answers: 1   Comments: 0

For a particle of a rotating rigid body, v = rω. So (1) ω ∝ (1/r) (2) ω ∝ v (3) v ∝ r (4) ω is independent of r

Foraparticleofarotatingrigidbody,v=rω.So(1)ω(1/r)(2)ωv(3)vr(4)ωisindependentofr

Question Number 25013    Answers: 0   Comments: 3

With reference to figure of a cube of edge a and mass m, state whether the following are true or false. (O is the centre of the cube.) (1) The moment of inertia of cube about z-axis is, I_z = I_x + I_y (2) The moment of inertia of cube about z′ is, I_(z′) = I_z + ((ma^2 )/2) (3) The moment of inertia of cube about z′′ is, I_(z′) = I_z + ((ma^2 )/2) (4) I_x = I_y

Withreferencetofigureofacubeofedgeaandmassm,statewhetherthefollowingaretrueorfalse.(Oisthecentreofthecube.)(1)Themomentofinertiaofcubeaboutzaxisis,Iz=Ix+Iy(2)Themomentofinertiaofcubeaboutzis,Iz=Iz+ma22(3)Themomentofinertiaofcubeaboutzis,Iz=Iz+ma22(4)Ix=Iy

Question Number 25000    Answers: 1   Comments: 0

find x,y from the equation: (1/2)x−yi+(1/(1+i))=((√(1+ω^8 ))+(√(1+ω^(10) )))^4

findx,yfromtheequation:12xyi+11+i=(1+ω8+1+ω10)4

Question Number 24997    Answers: 0   Comments: 0

Question Number 24985    Answers: 0   Comments: 0

Question Number 24961    Answers: 0   Comments: 0

Question Number 24945    Answers: 0   Comments: 4

A particle of mass m moving with a speed v hits elastically another stationary particle of mass 2m on a smooth horizontal circular tube of radius r. The time in which the next collision will take place is equal to

Aparticleofmassmmovingwithaspeedvhitselasticallyanotherstationaryparticleofmass2monasmoothhorizontalcirculartubeofradiusr.Thetimeinwhichthenextcollisionwilltakeplaceisequalto

Question Number 24934    Answers: 0   Comments: 3

A uniform circular disc of mass 1.5 kg and radius 0.5 m is initially at rest on a horizontal frictionless surface. Three forces of equal magnitude F = 0.5 N are applied simultaneously along the three sides of an equilateral triangle xyz with its vertices on the perimeter of the disc. One second after applying the forces, the angular speed of the disc in rad/s is :

Auniformcirculardiscofmass1.5kgandradius0.5misinitiallyatrestonahorizontalfrictionlesssurface.ThreeforcesofequalmagnitudeF=0.5Nareappliedsimultaneouslyalongthethreesidesofanequilateraltrianglexyzwithitsverticesontheperimeterofthedisc.Onesecondafterapplyingtheforces,theangularspeedofthediscinrad/sis:

Question Number 24909    Answers: 0   Comments: 0

Question Number 24893    Answers: 0   Comments: 2

About a collision which of the following are not correct a. Physical touch is a must b. Particles cannot change c. Effect of external force is not considered d. Momentum may or may not change multi−correct question

Aboutacollisionwhichofthefollowingarenotcorrecta.Physicaltouchisamustb.Particlescannotchangec.Effectofexternalforceisnotconsideredd.Momentummayormaynotchangemulticorrectquestion

Question Number 24879    Answers: 0   Comments: 4

A particle of mass m is fixed to one end of a light rigid rod of length l and rotated in a vertical circular path about its other end. The minimum speed of the particle at its highest point must be

Aparticleofmassmisfixedtooneendofalightrigidrodoflengthlandrotatedinaverticalcircularpathaboutitsotherend.Theminimumspeedoftheparticleatitshighestpointmustbe

Question Number 24877    Answers: 0   Comments: 7

Question Number 24858    Answers: 2   Comments: 0

If three positive numbers a, b, c are in A.P. and (1/a^2 ), (1/b^2 ), (1/c^2 ) also in A.P., then (1) a = b = c (2) 2b = 3a + c (3) b^2 = ((ac)/8) (4) 2c = 2b + a

Ifthreepositivenumbersa,b,careinA.P.and1a2,1b2,1c2alsoinA.P.,then(1)a=b=c(2)2b=3a+c(3)b2=ac8(4)2c=2b+a

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