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you have to see for what purpose you are calculating binding energy.if U want to disintegrate nucleaus U have to give energy to break it. If nucleus is forming Then energy is released.
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h=0.5xgxt^2 =5x36=180 now,180=4x v v = 45
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When external torque is zero, angular momentum of the system is conserved. it is analogous to the fact that linear momentum is conserved if external force is zero.
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For a thin convex lens when an object is placed at center of curvature of a biconvex lens (both radii same) , the image is formed at center of curvature on other side. Similarly all rays parrallel to principal axis pass through focus after refraction. All this has been tought through ray diagrams. The result has been proved grometrically also that for paraxial rays focal length f=R/2.
Now if I use the lens makers formula for same biconvex lens, I get f=R, Why not f=R/2 ? Where is all the mistake I am making.
Immediate help requested .
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FIRST UNDERSTAND WHY FIELD INSIDE CONDUCTORS IS ALWAYS ZERO.
CONDUCTORS BY DEFINITION CAN CARRY CHARGE THROUGH THEM. tHEY ARE HAVING FREE ELECTRONS IN ABUDANCE. THUS CHARGE SETS A CURRENT INSIDE THE CONDUCTOR AND WITHIN A SHORT TRANSITION TIME ALL THE CHARGE GO TO SURFACE, YOU CAN SAY BY REPULSION. SO WHEN THERE IS NO CHARGE INSIDE, THE FIELD IS ZERO. TAKE A GUASSIAN SURFACE JUST BELOW THE SURFACE OF YR CONDUCTOR, NO CHARGE INSIDE SO FIELD IS ZERO AS PER GAUSS LAW ALSO.
NOW WHEN CHARGE IS INSIDE ACAVITY IT INDUCES AN OPPSITE CHARGE AT INNER SURFACE AND SIMILAR CHARGE AT OUTER SURFACE. nOW IF YOU TAKE GUASSIAN SURFACE INSIDE BODY OF CUNDUCTOR NET CHARGE INSIDE GAUSSIAN SURFACE IS ZERO HENCE FIELD IS ZERO. IT IS DUE TO CAPABILITY OF CONDUCTORS TO GET INDUCED CHARGES WHICH IS DUE TO FREE CHARGE CARRIERS.
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(A)since j=I/A SO IF LENGTH IS DUOBLED AREAWILL BECOME HALF(vol constt). HENCE J=2j
(B)SINCE RESISTANCE IS PROPORTIONAL TO L / A SO RESISTANCE WILL INCREASE 4 TIMES. AS SUCH, CURRENT WILL BE ONE FOURTH. CONSEQUENTLY CURR DENSITY WILL HALF TIMES.
PLEASE RATE IF CORRECT
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use F= -dU/dx,
for small x,eliminate term 2b/x^2 , You should get F=-2ax
Note it is very similar to spring force F=-kx
so ang. freequency(w)=sqrt(2a/m)
T=2pi.Sqrt(2a/m).
If correct pl rate.
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Answer should be sqroot of 72=8.5
since total energy at C and D will be same
-2qqk/5 + 4 = -2qqk/root(x^2 + 9) +o
PE KE PE KE
at c at d
put k=9x10^9 and q=5x10^-5
=> x^2=72
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This question says that when a dielctric slab is just released inside parrallel plate capacitor whith a length a inside the slab will execute periodic motion. Find its time period. I am not able to relate the force on slab with its distance inside. Please explain and write the equation of motion.
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In a copper conductor of dia 2mm which has free electron density of 8.5x10**22 ,if a current of 2 Amp flows then drift velocity of ectrons is of the order of only 0.036 mm/s. This is extremely small speed. Now if Time rate of flow of charge is electric current then how elctricity apears to be traveling at very high speed, I mean when we switch on a bulb, even located at 100 m away it glows almost instantly. How this happens? Can we calculate speed of electric current if a 12 V battery is used in circuit of copper wire dia 2mm having a length of 100 mtr connecting a bulb of 5W. Take suitable value of resistance of copper wire? asusme any other data if required. Hats assured. quick answers prefered.
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Any body please provide solution of Question 57 HCV part II page 169.
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A solid sphere rolling on arough horizontal surface with a linear speed v collides elastically with a fixed, smooth, vertical wall. Find the speed of the sphere after it has started pure rolling in backward direction.
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A solid sphere rolling on arough horizontal surface with a linear speed v collides elastically with a fixed, smooth, vertical wall. Find the speed of the sphere after it has started pure rolling in backward direction.
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question appears incomplete or incorrect. If ground is frictionless then answer is 240 lb. if there is some friction on ground, check what is coefficient of friction!
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frame your question clearly with diagram or clear details. It is simple!
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produce Torque. Grip with two fingers generate enough friction force to produce required Torque. Single fingre can opnen if little torque is needed as little frictional force is possible.
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T is tension in string then, 2T=(30+60)g=900Newton =>T=450
See stability inside box, T and N are upward and wieght is downward so, N+T=60g=600Newton
N+450=600 =>N=150 Newton or 15 Kg
Now man wants that Machine should show a reading of 60 kg( machine indicates the normal reaction)
This time N is 600 but the important thing is that now system will not be static. it will under some acceleration upward to show higher normal reaction, man wants to see a reading of 600 new.! Let it is a and now tension is T' .
from outside the box 2T'-90g =90a
within box (no relative motion, but machine will experience pseudo force) so N'+T'-60a-60g=0 note N' will be 600 so
600 +T' =60a +60x10 =>T'=60 a
so 2T' -90x10 = 90xT'/60 => T'=1800 newton.
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Ans is (a) because P is at higher temp. so it should normally radiate at higher energy(smaller wavelength) .If its wavelength is of the erder of lower temp star Q then it must recede away faster to cause required red shift.
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(i)A uniform rod pivoted at its upper end hangs vertically. It is displaced through an angle of 60 degrees and then released. Find the magnitude of force acting on a particle of mass dm at the tip of the rod when the rod makes an angle of 37 degrees with the vertical. ( It is Q 69 ch10 HC Verma). Please help (ii) It is Q72. A string is wrapped over the edge of a uniform disc and the free end is fixed with ceiling. The disc moves down, unwinding the string. Find the downward acceleration of the disc.
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